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Elements of Mechanism 



PETER SCHWAMB, S.B. 

Professor of Machine Design (Retired), Massachusetts Institute of Technology 

ALLYNE L. MERRILL, S.B. 

Professor of Mechanism, Massachusetts Institute of Technology 

WALTER H. JAMES, S.B. 

Assistant Professor of Mechanical Engineering Drawing, 
Massachusetts Institute of Technology 



THIRD EDITION, REWRITTEN, ENLARGED, AND RESET \S 
TOTAL ISSUE, THIRTY THOUSAND 



NEW YORK 

JOHN WILEY & SONS, Inc. 

London: CHAPMAN & HALL, Limited 
1921 



CVj>vj 






Copyright, 1904, 1905, 

BY 

PETER SCHWAMB 

AND 

ALLYNE L. MERRILL 



Copyright, 1921, 

BY 

PETER SCHWAMB 

ALLYNE L. MERRILL 

WALTER H. JAMES 



/ 
JAN 24 1921 



ATLANTIC PRINTING COMPANY 
CAMBRIDGE, MASS., V. S. A 



©CI.A605488 l 






PREFACE TO THE THIRD EDITION 

This revised edition embodies many changes suggested by instructors 
who have used the second edition during the last sixteen years. The 
authors acknowledge their indebtedness to all of these gentlemen, and 
especially to Prof. George W. Swett, who has read a large part of the 
manuscript and given much valuable criticism. 

Professor Walter H. James, in charge of the^instruction in Mechanical 
Engineering Drawing and Machine Drawing at the Massachusetts 
Institute of Technology, and also an Instructor in Mechanism for many 
years, has been in charge of the revision, and joins in the authorship 
of the book. 

Peter Schwamb. 
Allyne L. Merrill. 
Walter H. James. 
Cambridge, Mass., 
December, 1920 



PREFACE TO SECOND EDITION 

The main subject-matter of this work was written during 1885 by- 
Peter Schwamb and has been used since then, in the form of printed 
notes, at the Massachusetts Institute of Technology, as a basis for 
instruction in mechanism, being followed by a study of the mechanism 
of machine tools and of cotton machinery. The notes were written 
because a suitable text-book could not be found which would enable 
the required instruction to be given in the time available. They have 
accomplished the desired result, and numerous inquiries have been 
received for copies from various institutions and individuals desiring to 
use them as text-books. This outside demand, coupled with a desire 
to revise the notes, making such changes and additions as experience 
has proved advisable, is the reason for publishing at this time. 

Very little claim is made as to originality of the subject-matter 
which has been so fully covered by previous writers. Such available 
matter has been used as appeared best to accomplish the object desired. 
Claim for consideration rests largely on the manner of presenting the 
subject, which we have endeavored to make systematic, clear, and 
practical. 

Among the works consulted and to which we are indebted for sug- 
gestions and illustrations are the following: "Kinematics of Machin- 
ery," and "Der Konstrukteur," by F. Reuleaux, the former for the 
discussion of linkages, and the latter for various illustrations of mechan- 
isms; "Principles of Mechanism," by S. W. Robinson, for the discus- 
sion of non-circular wheels; "Kinematics," by C. W. MacCord, for 
the discussion of annular wheels and screw-gearing; "Machinery and 
Millwork," by Rankine; "Elements of Mechanism," by T. M. Goodeve; 
and "Elements of Machine Design," by W. C. Unwin. 

Peter Schwamb. 
Allyne L. Merrill. 
October 20, 1904 



CONTENTS 

Page 
CHAPTER I 
Introduction 1 

CHAPTER II 
Revolving and Oscillating Bodies 12 

CHAPTER III 

Belt, Ropes and Chains 21 

CHAPTER IV 

Transmission op Motion by Bodies in Pure Rolling Contact 63 

CHAPTER V 
Gears and Gear Teeth 86 

CHAPTER VI 
Wheels in Trains 153 

CHAPTER VII 

Epicyclic Gear Trains 166 

CHAPTER VIII 
Inclined Plane, Wedge, Screw, Worm and Wheel 180 

CHAPTER X 

Four-Bar Linkage. Relative Velocities of Rigidly Connected Points 221 

CHAPTER XI 

LlNKWORK 244 

CHAPTER XII 

Straight-Line Mechanisms — Parallel Motions 291 

CHAPTER XIII 

Miscellaneous Mechanisms — Aggregate Combinations — Pulley 

Blocks — Intermittent Motion 306 

Problems 336 

Index 367 

vii 



Elements of Mechanism 



CHAPTER I 
INTRODUCTION 

1. A Machine is a combination of resistant bodies so arranged 
that by their means the mechanical forces of nature can be compelled 
to produce some effect or work accompanied with certain determinate 
motions.* In general, it may be properly said that a machine is an 
assemblage of parts interposed between the source of power and the 
work, for the purpose of adapting the one to the other. Each of the 
pieces in a machine either moves or helps to guide some of the other 
pieces in their motion. 

No machine can move itself, nor can it create motive power; this 
must be derived from external sources, such as the force of gravita- 
tion, the uncoiling of a spring, or the expansion of steam. As an 
example of a machine commonly met with, an engine might be men- 
tioned. It is able to do certain definite work, provided some external 
force shall act upon it, setting the working parts in motion. It con- 
sists of a fixed frame, supporting the moving parts, some of which 
cause the rotation of the engine shaft, others move the valves distrib- 
uting the steam to the cylinder, and still others operate the governor 
which controls the engine. These moving parts are so arranged that 
they make certain definite motions relative to each other when an 
external force, as steam pressure, is applied to the piston. 

2. A Mechanism is a portion of a machine where two or more 
pieces are combined, so that the motion of the first compels the 
motion of the others, according to a law depending on the nature of 
the combination. For example the combination of a crank and con- 
necting-rod with guides and frame, in a steam engine, serving to con- 
vert reciprocating into circular motion, would thus be called a 
mechanism. 

The term Elementary Combination is sometimes used synonymously 
with A Mechanism. 

A machine is made up of a series or train of mechanisms. 
* Reuleaux: Kinematics of Machinery. 
1 



2 ELEMENTS OF MECHANISM 

3. The Science of Mechanism treats of the forms of the pieces 
used in the construction of machinery and the laws which govern the 
motions and forces existing in or transmitted by them. 

The operation of any machine depends upon two things: first, the 
transmission of certain forces, and second, the production of deter- 
minate motions. In designing, due consideration must be given to 
both of these, so that each part may be adapted to bear the stresses 
imposed on it, as well as have the proper motion relative to the other 
parts of the machine. The nature of the movements does not depend 
upon the strength or absolute dimensions of the moving parts, as can 
be shown by models whose dimensions may vary much from those 
requisite for strength, and yet the motions of the parts will be the same 
as those of the machine. Therefore, the force and the motion may be 
considered separately, thus dividing the science of Mechanism into 
two parts, viz.: 

1° Pure Mechanism, which treats of the motion and forms of the 
parts of a machine, and the manner of supporting and guiding them, 
independent of their strength. 

2° Constructive Mechanism, which involves the calculation of the 
forces acting on different parts of the machine; the selection of ma- 
terials as to strength and durability in order to withstand these 
forces, taking into account the convenience for repairs, and facilities 
for manufacture. 

What follows, will, in general, be confined to the first part, pure 
mechanism, or what is sometimes called "the geometry of machinery "■; 
but in some cases the forces in action will be considered. 

The definition of a machine might be modified to accord with the 
above, as follows: 

A Machine is an assemblage of moving parts so connected that 
when the first, or recipient, has a certain motion, the parts where the 
work is done, or effect produced, will have certain other definite 
motions. 

4. Driver and Follower. That piece of a mechanism which is 
supposed to cause motion is called the driver, and the one whose 
motion is effected is called the follower. 

5. Frame. The frame of a machine is a structure that supports 
the moving parts and regulates the path, or kind of motion, of many 
of them directly. In discussing the motions of the moving parts, it 
is convenient to refer them to the frame, even though it may have, 
as in the locomotive, a motion of its own. 

6. Modes of Transmission. If the action of natural forces of 
attraction and repulsion is not considered, one piece cannot move 



INTRODUCTION 3 

another, unless the two are in contact or are connected to each other 
by some intervening body that is capable of communicating the 
motion of the one to the other. In the latter case, the motion of the 
connector is usually unimportant, as the action of the combination 
as a whole depends upon the relative motion of the connected pieces. 
Thus motion can be transmitted from driver to follower: 

1° By direct contact. 

2° By intermediate connectors. 

7. Links and Bands. An intermediate connector can be rigid or 
flexible. When rigid it is called a link, and it can either push or pull, 
such as the connecting-rod of a steam engine. Pivots or other joints 
are necessary to connect the link to the driver and follower. 

If the connector is flexible, it is called a band, which is supposed to 
be inextensible, and only capable of transmitting a pull. A fluid 
confined in a suitable receptacle may also serve as a connector, as in 
the hydraulic press. The fluid might be called a pressure-organ in 
distinction from the band, which is a tension-organ. 

8. Motion is change of position. Motion and rest are necessarily 
relative terms within the limits of our knowledge. We may conceive 
a body as fixed in space, but we cannot know that there is one so 
fixed. If two bodies, both moving in space, remain in the same posi- 
tion relative to each other, they are said to be at rest, one relatively 
to the other; if they do not, either may be said to be in motion rela- 
tively to the other. 

Motion may thus be either relative, or it may be absolute, provided 
some point is assumed as fixed. In what follows, the earth will be 
assumed to be at rest, and all motions referred to it will be considered 
as absolute. 

9. Path. A point moving in space describes a line called its path, 
which may be rectilinear or curvilinear. The motion of a body is 
determined by the paths of three of its points not on a straight line. 
If the motion is in a plane, two points suffice, and if rectilinear, one 
point suffices to determine the motion. 

10. Direction. If a point is moving along a straight path the 
direction of its motion is along the line which constitutes its path; 
motion towards one end of the line being assumed as having positive 
direction and indicated by a -f- sign, the motion toward the other 
end would be negative and indicated by a — sign. If a point is 
moving along a curved path, the direction at any instant is along the 
tangent to the curve and may be indicated as positive or negative, 
as in the case of rectilinear motion. 



4 ELEMENTS OF MECHANISM 

11. Continuous Motion. When a point continues to move indefi- 
nitely in a given path in the same direction, its motion is said to be 
continuous. In this case the path must return on itself, as a circle 
or other closed curve. A wheel turning on its bearings affords an 
example of this motion. 

12. Reciprocating Motion. When a point traverses the same path 
and reverses its motion at the ends of such path, the motion is said 
to be reciprocating. 

13. Vibration and Oscillation are terms applied to reciprocating 
circular motion, as that of a pendulum. 

14. Intermittent Motion. When the motion of a point is inter- 
rupted by periods of rest, its motion is said to be intermittent. 

15. Revolution and Rotation. A point is said to revolve about an 
axis when it describes a circle of which the center is in, and the plane 
is perpendicular to, that axis. When all the points of a body thus 
move, the] body is said to revolve about the axis. If this axis passes 
through the body, as in the case of a wheel, the word rotation is used 
synonymously with revolution. The word turn is often used synony- 
mously with revolution and rotation. It frequently occurs that a body 
not only rotates about an axis passing through itself, but also moves in 
an orbit about another axis. 

16. An Axis of Rotation or revolution is a line whose direction is not 
changed by the rotation ; a fixed axis is one whose position, as well as 
its direction, remains unchanged. 

17. A Plane of Rotation or revolution is a plane perpendicular to 
the axis of rotation or revolution. 

18. Direction of Rotation or revolution is defined by giving the direc- 
tion of the axis and stating whether the turning is Right-handed 
(clockwise) or Left-handed (anti-clockwise). 

19. Cycle of Motions. When a mechanism is set in motion and its 
parts go through a series of movements which are repeated over and 
over, the relations between and order of the different divisions of the 
series being the same for each repetition, one of these series is called 
a cycle of motions. For example, one revolution of the crank of a 
steam engine causes a series of different positions of the piston-rod, 
and this series of positions is repeated over and over for each revolu- 
tion of the crank. 

20. Period of motion is the time occupied in completing one cycle. 

21. Linear Speed is the rate of motion of a point along its path, or 
the rate at which a point is approaching or receding from another 
point in its path. If the point to which the motion of the moving 
point is referred, is fixed, the speed is the absolute speed of the point. 



INTRODUCTION , 5 

If the reference point is itself in motion the speed of the point in 
question is relative. Linear speed is expressed in linear units, per 
unit of time. 

22. Angular Speed is the rate of turning of a body about an axis, 
or the rate at which a line on a revolving body is changing direction, 
and is expressed in angular units per unit of time. 

In case a body is revolving about an axis outside of itself, any 
point in the body has only linear speed, but a line, real or imaginary, 
joining the point to the axis of revolution has angular speed, also a line 
joining any two points on the body has angular speed. 

23. Uniform and Variable Speed. Speed is uniform when equal 
spaces are passed over in equal times, however small the intervals 
into which the time is divided. The speed in this case is the space 
passed over in a unit of time, and if s represent the space passed over 
in the time t, the speed v will be 

-r <» 

Speed is variable when unequal spaces are passed over in equal 
intervals of time, increasing spaces giving accelerated motion and 
decreasing spaces giving retarded motion. The speed, when variable, 
is the limit of the space passed over in a small interval of time, divided 
by the time, when these intervals of time become infinitely small. If 
s represents the space passed over in the time t, then 

v = limit of — as At approaches zero, 

ds ,_. 

v = Ti - (2) 

The uniform speed of a point or line is measured by the number of 
units of distance passed over in a unit of time, as feet per minute, 
radians per second, etc. When the speed is variable it is measured 
by the distance which would be passed over in a unit of time, if the 
point or line retained throughout that time the speed which it had at 
the instant considered. 

24. Velocity is a word often used synonymously with speed, al- 
though, accurately speaking, velocity includes direction as well as 
speed. The linear velocity of a point is not fully defined unless the 
direction in which it is moving and the rate at which it is moving are 
both known. The angular velocity of a line would be defined by 
giving its angular speed and the direction of the perpendicular to the 
plane in which the line is turning. 

25. Linear Acceleration is the rate of change of linear velocity. 
Since velocity involves direction as well as rate of motion, linear 



6 ELEMENTS OF MECHANISM 

acceleration may involve a change in speed or direction, or both. 
Any change in the speed takes place in a direction tangent to the 
path of the point and is called tangential acceleration, while a change 
in direction takes place normal to the path and is called normal 
acceleration. This must never be confused with angular acceleration 
which will be discussed later. In general in this text only tangential 
acceleration will be considered and it will be understood that when 
the word acceleration is used in connection with linear motion it is 
intended to refer to the tangential acceleration unless normal accelera- 
tion is definitely mentioned. The following example will make clear 
the meaning of terms in which tangential acceleration is expressed: 

If a body is moving at the rate of 1 ft. per second at the end of the 
first second, 3 ft. per second at the end of the second second, 5 ft. 
per second at the end of the third second, etc., the speed is increasing 
at the rate of 2 ft. per second each second. Its acceleration is two feet 
per second each second. 

Acceleration , may be either -positive or negative. If the speed is 
increasing the acceleration is positive; if the speed is decreasing the 
acceleration is negative. If the speed changes the same amount 
each second the acceleration is uniform, but if the speed changes by 
different amounts at different times the acceleration is variable. If 
Av represents the change in speed in the time At, then, if the accelera- 
tion a is uniform 

Ay 



<« 



When the acceleration is variable 



a = limit of — as At approaches zero 

dv ,.>. 

a = Tf (4) 

26. Angular Acceleration is rate of change of angular velocity. In 
this case, as in the case of linear acceleration, a change in either speed 
or direction of rotation, or both, may be involved. For example, if 
a line is turning in a plane with a varying angular speed it has angu- 
lar acceleration which may be positive or negative; or if the direction 
of the plane of rotation is changing the line also has angular accelera- 
tion. Unless otherwise stated angular acceleration in this text will 
be understood to refer to change in angular speed. Angular accelera- 
tion is expressed in angular units change in speed per unit time (such 
as radians, degrees, or revolutions per minute each minute). Equa- 
tions (3) and (4) will apply to angular acceleration if a and v are 
expressed in angular units. 



Variable motion 



INTRODUCTION / 

27. Kinds of Motion. From the preceding discussion it is evident 
that motion, whether absolute or relative, may be classified as follows : 

Uniform motion — Acceleration zero. 

Acceleration constant 
Acceleration variable 
Acceleration constant part of the 
time and variable part of the time. 

A body having uniform motion travels without change of speed, 
if any normal acceleration is neglected. For example, if a block is 
sliding on guides a distance of one inch each second, or a wheel turn- 
ing constantly at the rate of 100 revolutions per minute, each would 
be said to have uniform motion. 

The most familiar example of motion having constant acceleration 
is that of a body falling freely under the action of gravity. A weight 
dropped from a height will have at the start a zero speed. Under 
the action of the constant force of gravity its speed will increase at a 
constant rate such that at the end of one second the body will be 
moving at the rate of about 32.2 ft. per second and it will have 
dropped 16.1 ft. At the end of two seconds it will have a speed of 
64.4 ft. per second and will have dropped 48.3 ft. If the constant 
force acting were of different magnitude the amount of acceleration 
would be different but it would still be constant. The space moved 
over in successive intervals of time by a body having constant accelera- 
tion will be in the ratio of the odd numbers, 1, 3, 5, 7, 9, etc. 

In the case of machine parts having variable acceleration the 
variation does not usually follow any simple law. 
One form of motion with variable acceleration, 
however, occurs not infrequently, and is known as 
simple harmonic motion. If a point, as R, Fig. 1, 
moves with uniform speed around the circumference FlG 1 

of a semicircle and another point T moves across 
the diameter in the same length of time, the speed of T varying so 
that it will always be at the foot of a perpendicular let fall from R to 
the line A B, then T is said to have harmonic motion. 

The relation between the displacement AT of the point T from its 
initial position and the angle is expressed by the equation 

AT = OA - OR cos = OR (1 - cos 0). (5) 

The acceleration of the point T may be deduced as follows : 

Let s = AT; co = angular speed of OR in radians per second; t = time 
required for line OR to move through angle 0, therefore = ut ■ v = speed 




8 ELEMENTS OF MECHANISM 

of T in linear units per second, a = acceleration in similar units per 

second each second. 

Then from equation (5) 

s = OR - OR cos 6 = 0R -OR cos ut. 

ds 
From equation (2) v = -r = uOR sin cat, 





dv d-s n 
a = -rr = -rr* = co 2 OR cos cot, 
dt dt 2 


but 


x OT 

C0S " t = 6R- 


Therefore 


OT 

a = u-OR X ^=r = co 2 07\ 



(5a) 

Therefore the acceleration is proportional to the distance of the. mov- 
ing point from the center of its path. When T is approaching its 
acceleration is positive and when receding from the acceleration is 
negative. 

The arrangement of parts in a mechanism may be such that one or 
more of the pieces has an increasing speed at the beginning, then moves 
at uniform speed during the greater part of the motion, and has de- 
creasing speed at the end. 

28. Modification of Motion. In the action of a mechanism the 
motion of the follower may differ from that of the driver in kind, in 
speed, in direction, or in all three. As the paths of motion of the 
driver and follower depend upon the connections with the frame of the 
machine, the change of motion in kind is fixed, and it only remains to 
determine the relations of direction and speed throughout the motion. 
The laws governing the changes in direction and speed can be deter- 
mined by comparing the movements of the two pieces at each instant 
of their action, and the mode of action will fix the laws. Therefore, 
whatever the nature of the combination, if it is possible to determine, 

throughout the motion of the 
driver and follower, the speed 
ratio and directional relation, 
the analysis will be complete. 





Wssg^ssp^^^^ .gP' Either the speed ratio or the 

A ) l \ J- + ) \ _ J ^jP^" directional relation may vary, 

or remain the same throughout 

the action of the two pieces. 

FlG 2 29. Pairs of Elements. In 

order that a moving body, as A 

(Fig. 2), may remain continually in contact with another body B, and 

at the same time move in a definite path, B would have a shape which 



INTRODUCTION 9 

could be found by allowing A to occupy a series of consecutive positions 
relative to B, and drawing the envelope of all these positions. Thus, 
if A were of the form shown in the figure, the form of B would be that 
of a curved channel. Therefore, in order to compel a body to move in 
a definite path, it must be paired with another, the shape of which is 
determined by the nature of the relative motion of the two bodies. 

30. Closed or Lower Pair. If one element not only forms the 
envelope of the other, but encloses it, the forms of the elements being 
geometrically identical, the one being solid or full, and the other 
being hollow or 7 open, we have what may be called a closed pair, also 
called a lower pair. In such a pair, surface contact exists between the 
two members. 

On the surfaces of two bodies forming a closed pair, coincident lines 
may be supposed to be. drawn, one on each surface; and if these lines 
are of such form as to allow them to move along each other, that is, allow 
a certain motion of the two bodies paired, three forms only can exist : 

1° A straight line, which allows straight translation (Fig. 3). 

2° Among plane curves, or curves of two dimensions, a circle, 
which allows rotation, or revolution (Figs. 4 and 5). 

3° Among curves of three dimensions, the helix, which allows a 
combination of rotation and straight translation (Fig. 6). 



I Liz 



i „ . , i 



\ 



Fig. 3 




Fig. 5 




31. Higher Pairs. The pair represented in Fig. 2 is not closed, as 
the elementary bodies A and B do not enclose each other in the above 
sense. Such a pair is called a higher pair and the contact between the 
elements is along Unes only. 

32. Incomplete Pairs of Elements. Hitherto it has been assumed 
that the reciprocal restraint of two elements forming a pair was com- 
plete, i.e., that each of the two bodies, by the rigidity of its material 



10 ELEMENTS OF MECHANISM 

and the form given to it, restrained the other. In certain cases it is 
only necessary to prevent forces having a certain definite direction 
from affecting the pair, and in such cases it is no longer absolutely 
necessary to make the pair complete; one element can then be cut 
away where it is not needed to resist the forces. 

The bearings for railway axles, the steps for water-wheel shafts, the 
ways of a planer, railway wheels kept in contact with the rails by the 
force of gravity are all examples of incomplete pairs in which the ele- 
ments are kept in contact by external forces. 

33. Inversion of Pairs. In Fig. 3 if B is the fixed piece all points 
on A move in straight lines. If A were the fixed piece all points in B 
would move in straight lines. That is, the absolute motion of the 
moving piece is the same, whichever piece is fixed. The same state- 
ment holds true of the pairs shown in Figs. 4, 5 and 6. 

This exchange of the fixedness of an element with its partner is 
called the inversion of the pair, and in the case of any closed or lower 
pair it does not effect either the absolute or the relative motion. 

In the pairs shown in Figs. 2 and 7, both of which are higher pairs, 
the relative motion of A and B is the same when A is fixed as when B is 
fixed. The absolute motion of A when B is fixed is 
not the same as the absolute motion of B when A is 
fixed. 
i 34. Bearings. The word bearing is applied, in 
F - general, to the surfaces of contact between two 

pieces which have relative motion, one of which 
supports or partially supports the other. One of the pieces may be 
stationary, in which case the bearing may be called a stationary 
bearing; or both pieces may be moving. 

The bearings may be arranged, according to the relative motions 
they will allow, in three classes: 

1° For straight translation the bearings must have plane or cylin- 
drical surfaces, cylindrical being understood in its most general 
sense. If one piece is fixed the surfaces of the moving pieces are called 
slides; those of the fixed pieces, slides or guides. 

2° For rotation, or turning, the bearings must have surfaces of 
revolution, as circular cylinders, cones, conoids or flat disks. The 
surface of the solid or full piece is called a journal, neck, spindle or 
pivot; that of the hollow or open piece, a bearing, gudgeon, pedestal, 
plumber-block, pillow-block, bush or step. 

3° For translation and rotation combined, or helical motion, they 
must have a helical or screw shape. Here the full piece is called a 
screw and the open piece a nut. 




INTRODUCTION 



11 



35. Collars and Keys. It is very often the case that pulleys or 
wheels are to turn freely on their cylindrical shafts and at the same 
time have no motion along them. For this purpose, rings or collars 
(Fig. 8a) are used; the collars D and E, held by set screws, prevent 
the motion of the pulley along the shaft but allow it free rotation. 
Sometimes pulleys or couplings must be free to slide along their 




shafts, but at the same time must turn with them; they must then 
be changed to a sliding pair. This is often done by fitting to the 
shaft and pulley or sliding piece a key C (Fig. 8b), parallel to the 
axis of the shaft. The key may be made fast to either piece, the 
other having a groove in which it can slide freely. The above ar- 
rangement is very common, and is called a feather and groove or 
spline, or a key and keyway. 



CHAPTER II 
REVOLVING AND OSCILLATING BODIES 



36. Revolving Bodies.* One of the most common motions in ma- 
chinery is revolution or rotation about an axis. The rotating body 
may be a cylinder, or cone, or a piece of irregular form. The shaft or 
element upon which the rotating piece is supported may turn with it, 
being itself supported in bearings and restrained from moving endwise 
by collars; or the shaft may be held stationary and the piece turn on 
the shaft. An example of the first case is shown in Fig. 9 and of the 
second case in Fig. 10. 

37. Angular Speed. Suppose some force is applied to the shaft in 
Fig. 9 so that it is caused to turn around, say, 75 times in a minute. 



Colla 





Fig. 9 



Fig. 10 



If the wheel is fast to the shaft, the wheel will turn around 75 times 
in a minute. It would be said, then, that the wheel makes 75 revolu- 
tions per minute (usually abbreviated thus: 75 r.p.m.) It is common 
practice to speak of the angular speed or speed of turning in terms of 
revolutions per unit of time, usually per minute or second. 

Another unit sometimes used for measuring angular motion is the 
angle called the radian. This is the angle which is subtended by the 
arc of a circle equal in length to the radius. Since the radius is con- 
tained in the circumference of a circle 2 t times, there must be 2 w 
radians in 360°, or one radian is equal to 360° -f- 2 t = 57° 17' 42". 

If a revolving wheel turns once per minute, its angular speed as we 
have already seen is 1 r.p.m., and since one revolution of the wheel causes 

* Throughout this book the words revolve and turn will refer to turning about any 
axis, whether within or outside the body in question, while rotate will refer only to 
turning about an axis passing through the body. 

12 



REVOLVING AND OSCILLATING BODIES 13 

any radial line on the wheel to sweep over 360° or 2 t radians, the angu- 
lar speed of the wheel is 2 t radians per minute. Now, if the wheel 
turns N times per minute, the angular speed is N r.p.m. or 2 tN radians 
per minute. That is, 

Angular speed in radians = 2tX number of revolutions. (6) 

Example 1. If a wheel turns 90 r.p.m. its angular speed in radians is 
2-7T 90 = 565.5 radians per minute. 

Referring to Fig. 11, let the body M be rigidly attached to an arm 
which is turning around the axis C, the arm and M revolving together. 
Then the lines CA and CB which join any two points A and B to the 
axis have angular speed about C and since the entire body is rigid and 
the angle ACB is constant, CA and CB each have the same angular 
speed as the arm. Moreover, since, as the body revolves, the line AB 
constantly changes direction, it may also be said to have angular speed, 
which, in this case, is the same as that of the lines CA and CB. 

If M is not rigidly attached to the arm but is rotating on the axis S 
which is carried by the arm, as in Fig. 12, the lines CA, CB and AB will 




Fig. 11 Fig. 12 

no longer necessarily have the same angular speed since the angles 
turned through in a given time by these lines depend not only on the 
speed at which the arm is turning about C but also upon the speed at 
which M is turning about the axis S. 

38. Linear Speed of a Point on a Revolving Body. Consider a 
point A on the circumference of the wheel in Fig. 9. While the wheel 
turns once, A travels over the circumference of a circle of the same 
diameter as the wheel, or it travels 2 wR in. if the radius of the wheel 
is R in. Then, if the wheel turns N times in a unit of time, A travels 
over the circle N times in a unit of time. Therefore, the linear speed 
of the point A is 2 irRN in. per unit of time. Writing this in the form 
of an equation, 

Linear speed of A = 2 wRN. (7) 



14 



ELEMENTS OF MECHANISM 



Example 2. Suppose the wheel is 12 in. in diameter and turns 40 times per 
minute. The speed of A would be 

2 7r 6 X 40 in. per minute = 1506 in. per minute. 

From equation (6) it appears that the angular speed is equal to 2 ttN 
radians; so that, dividing equation (7) by equation (6) 

Linear speed of A 2tRN _ R 

Angular speed of wheel in radians 2 tN 1 
or, 

Linear speed of a point on a revolving body = angular speed of body 

in radians X distance of the point from the center. (8) 

39. Speed Ratio of Points at Different Distances from Axis. If 

another point B is chosen on the side of the wheel at a distance Ri 
from the center, it can be shown in the same way that Eq. (7) was 
derived, that 

Linear speed of B = 2-irRiN. (9) 

Now, dividing Eq. (7) by Eq. (9), 

Linear speed of A _ 2 tRN 
Linear speed of B 2 7r.KiiV 

Linear speed of A R 



Therefore, 



Linear speed of B Ri 



(10) 



Thus, the linear speeds of two points on a revolving wheel are directly 
proportional to the distances of the points from the center about which the 
wheel is turning. 

The linear speed of a point on the circumference of a wheel is often 
spoken of as the periphery-speed or surface speed of the wheel. 

Take another case, that of two wheels 
fast to the same shaft as shown in Fig. 13. 
The weight P is supposed to be hung from a 
steel band which is wound on the outside 
of wheel A and the weight W from another 
steel band wound on the outside of wheel B. 
Suppose that the shaft starts to turn in the 
direction shown by the arrow. Then the 
band which supports P will be paid out, 
that is, will unwind, at a speed equal to the 
periphery speed of A and the weight P will descend at that speed. 
At the same time the other band will be winding on to the wheel B 
and the weight W will be rising at a speed equal to the periphery speed 
of B. If N represents the number of turns per second of the shaft, 




Fig. 13 



REVOLVING AND OSCILLATING BODIES 15 

R the radius of A, Ri the radius of B, then the speed of P = 2 irRN 
and speed of W = 2 irRiN, or 

Speed P R 

Speed W Ri { } 

which is the same, equation found when both points were on the same 
wheel. • « 

Example 3. Let the diameter of A = 12 in. and the diameter of B = 8 in. and 
let the shaft turn If times per second. Then 

Speed P = 27rX6Xl| = 56.55 in. per second. 
Speed Tf=27rX4Xl| = 37.70 in. per second. 

Now, according to Eq. 11, these speeds should be in the ratio of 6 to 4 or 1.5 to 1, 
and if 56.55 is divided by 37.7, the result is equal to 1.5 except for the slight error 
due to carrying the figures to the nearest hundredth of an inch. 

40. Relation between Forces and Speeds. Suppose that in Fig. 13 
it is assumed that there is no friction and that the weights of P and 
W are such that, if the shaft is at rest, the weights will just balance 
each other, or, if the shaft is caused to start turning in a given direction, 
the weights will allow it to keep turning at a uniform speed. 

The work done by a force is equal to the force, expressed in units 
of force, multiplied by the distance through which the force acts, ex- 
pressed in linear units, provided the motion takes place in the direc- 
tion in which the force acts. Now, if friction is neglected, the work 
obtained from a machine must equal the work put into it. Hence, in 
Fig. 13, if the falling weight P be considered as the force driving the 
machine, the work put into the machine is the weight P multiplied by 
the distance P falls. The work obtained from the machine is the 
weight of W multiplied by the distance W is raised. 

Then the weight of P (in pounds or other weight units) multiplied 
by the distance P moves in a given length of time is equal to the weight 
of W multiplied by the distance it moves in the same time, the units 
of weight and distance being the same in both cases. If the shaft is 
assumed to make N turns, the distances moved by P and W are 2 irRN 
and 2 ttRiN, respectively. 

Therefore, P X 2 tRN = W X 2 irRiN, (12) 

Weight of P fix 
Weight of W R K J 

Since from Eq. (11) the 

Linear speed of P _ R_ 
Linear speed of W Ri 



16 



ELEMENTS OF MECHANISM 



Therefore, 



(14) 



41. 

way i 



Weight of P = Speed of W 
Weight of W ~ Speed of P 
Cranks and Levers. A crank may be denned in a genei 
s an arm rotating or oscillating about an axis. It may bf 
thought of as a piece cut out oi 
a wheel or disk, as suggested by 
the dotted circle in Fig. 14, and 
the laws for revolving wheels 
apply equally to cranks. When 
two cranks are rigidly connected 
to each other the name lever is 
often applied to the combination, 
particularly when the motion is 
oscillating over a relatively small 
angle. 
In Fig. 15 the two arms of the lever are shown at an angle of 180° 
with each other. This condition does not necessarily hold, however, 
for the two arms may make any angle with each other from 180° as in 
Fig. 15 down to 0° as in Fig. 16. When the angle between the two 
arms is less than 90° as in Figs. 16 and 17 it is often called a bell crank 

U- 




vj_^ 



Fig. 14 



/H R- 



-4*— tf-r^! \ 



m 



\ A 



^ 



C ' B.i 




Fig. 15 

lever, and when the angle is more than 90° as in Figs. 15 and 18 it is 
often called a rocker. These terms, however, are used rather loosely 
and somewhat interchangeably. 

In all these cases the following equation holds true: 
Linear speed A _ Distance of A from axis 
Linear speed B Distance of B from axis 
* It must be always borne in mind that any equation such as Eq. (14) does 
not take friction into account. 



(15) 



REVOLVING AND OSCILLATING BODIES 



17 



The two lever arms may be in the same plane as in Figs. 15 to 18 or 
they may be attached to the same shaft but lie in different planes as 
li Fig. 19. 

f ^42. Motion from Levers. It is often necessary to transfer some 
small motion from one line to another. Three cases will be considered 
which depend on the relative positions of the lines of motion: 

1° Parallel lines. 

2° Intersecting lines. 

3° Lines neither parallel nor intersecting. 

The first case is an application of the form of lever shown in Figs. 15 
and 16, and also Fig. 19 if the arms CA and CB are in the same plane 

Fig. 18 





passing through the axis of the shaft. The motions of A and B are 
directly proportional to their distances from the axis C. In Fig. 15, 




Fig. 20a 



A and B are always moving in opposite directions, while, in Fig. 16, 
A and B move in the same direction. 

In the second case a bent lever is used of the type shown in Figs. 
17 and 18. Referring to Fig. 20a, let it be assumed that a lever is to 



18 



ELEMENTS OF MECHANISM 



be laid out which will give motion along AD bearing a known ratio to 
that along BD. 

Draw the line DC, dividing the angle ADB into two angles ADE 
and BDE whose sines are directly proportional to the motions required 
along AD and BD respectively. 

This may be done by erecting perpendiculars MN and PT on AD 
and BD in the ratio of the required motions along those lines, and draw- 
ing through their extremities N and T lines parallel to AD and BD 
respectively; the intersection of these lines at E determines the line 




Fig. 20b 

DE. Choose any point C in DE, and drop the perpendiculars CA 
and CB on AD and BD respectively; then ACB is the bell-crank lever 
required. As the lever moves 

Linear speed of A _ AC _ sin CDA 
Linear speed of B BC sin CDB 

It is evident that, for a small angular motion, the movements in 
AD and BD are very nearly rectilinear, and they will become more and 
more so the farther C is removed from the point D. 

Any slight motion that may occur perpendicular to the lines AD 
and BD may be provided for by the connectors used. It is to be 
noticed, however, that for a given motion on the lines AD and BD 
these perpendicular movements, or deviations, are less when the lever- 
arms vibrate equal angles each side of the positions which they occupy 
when perpendicular to the lines of motion, and they should always be 



REVOLVING AND OSCILLATING BODIES 



19 



arranged to so vibrate. By moving the point C nearer to D, at the 
same time keeping the lever-arms the same, this perpendicular devia- 
tion could be disposed equally on each side of the lines of motion 
which is advisable especially in cases where the deviation is allowed 




Fig. 21 

for by the spring of the connecting piece. This is shown in position 
AiCiBi, Fig. 20b. In Fig. 20 it will be seen that A and B move in 
opposite directions, while Fig. 21 shows the result if A and B are to 
move in the same direction. 




Fig. 22 



In the third case a lever of the type of Fig. 19 would be used with 
the arms CA and CB making the proper angle with each other. Let 



20 



ELEMENTS OF MECHANISM 



BD be the line along which B is to give motion and AD the line along 
which A is to give motion. Let BD lie in the plane XY and AD lie in 
the plane VW. To find the position of the line DC which is the trace 
of the plane containing the axis of the shaft, assume the plane VW to 
be moved to the left until it coincides with XY. Then lay out the 

lever in the left elevation as described 
for Fig. 20. Next assume the plane 
VW moved back to its proper posi- 
tion, carrying the arm CA with it. 

43. Effective Lever Arms. In 
the case of a lever in a position 





Fig. 23 



Fig. 24 



such as indicated in Figs. 23 or 24, the effect is the same, for the 
instant, as if the weights P and Pi were attached to the lever MCN, 
whose arms are found by drawing perpendiculars from the axis C to 
the line of action of the forces exerted by the weights P and P±. 
The perpendiculars CM and CN may be called the effective lever 
arms or moment arms of the weights. 



CHAPTER III 



BELTS, ROPES AND CHAINS 




44. Flexible Connectors. If the wheel A, Fig. 25, is turning at a 
certain angular speed about the axis S its outer surface will have a linear 
speed dependent upon the angular speed and the diameter of A. (See 
§38.) 

If a flexible band is stretched over A, connecting it with another 
wheel B and there is sufficient friction between the band and the sur- 
faces of the wheels to prevent 
appreciable slipping, then the 
band will move with a linear 
speed approximately equal to 
the surface speed of A, and will 
impart approximately the same 
linear speed to the surface of B, 
thus causing B to turn. The 
wheels may be on axes which 
are parallel, intersecting, or 
neither parallel nor intersecting. 
into three general classes: 

1° Belts made of leather, rubber, or woven fabrics are flat and 
thin, and require pulleys nearly cylindrical with smooth surfaces. 
Flat ropes may be classed as belts. 

2° Cords made of catgut, leather, hemp, cotton or wire are nearly 
circular in section and require either grooved pulleys or drums with 
flanges. Rope gearing, either by cotton or wire ropes, may be placed 
under this head. 

3° Chains are composed of links or bars, usually metallic, jointed 
together, and require wheels or drums either grooved, notched, or 
toothed, so as to fit the links of the chain. 

For convenience the word band may be used as a general term to 
denote all kinds of flexible connectors. 

Bands for communicating continuous motion are endless. 

Bands for communicating reciprocating motion are usually made 
fast at their ends to the pulleys or drums which they connect. 

21 



Flexible connectors may be divided 




22 ELEMENTS OF MECHANISM 

45. Pitch Surface and Line of Connection. Fig. 26 represents the 
edge view of a piece of a belt before being wrapped around the pulley. 
If it is assumed that there are no irregularities in the make up of the 
belt the upper surface o is parallel to and equal in length to the sur- 
face i. When this same belt is stretched around a pulley, as in Fig. 
27, the surface i is drawn firmly against the surface of the pulley while 
the surface o bends over a circle whose radius is greater than that of 
the surface of the pulley by an amount equal to the belt thickness 2 p. 

o The outer part of the belt must 

therefore stretch somewhat and 

/" ^ ((( \jf\ ' the inner part compress. There 

will be some section between i 

and o which is neither stretched 

_, _. ~ ~ nor compressed and the name 

Fig. 26 Fig. 27 , 7 • i 

neutral section may be given to 

this part of the belt. In the case of a flat belt the neutral section 

may be assumed to be half way between the outer and inner surfaces. 

An imaginary cylindrical surface around the pulley, to which the 

neutral section of the belt is tangent, is the pitch surface of the 

pulley, the radius of this being the effective radius of the pulley. 

A line in the neutral section of the belt at the center of its width is 

the line of connection between two pulleys and is tangent to the pitch 

surfaces, and coincides with a line in each pitch surface known as the 

pitch line. 

46. Speed Ratio and Directional Relation of Shafts Connected by a 
Belt. In Fig. 25 let the diameter of the pulley A be D inches, the di- 
ameter of B be Di inches and the half thickness of belt = p. Also let 
N represent the r.p.m. of S, and iVi = r.p.m. of Si. 

Then, from equation (7), 

Linear speed of pitch surface of A = tN (D + 2 p), 
and Linear speed of pitch surface of B = wNi (Di + 2 p) . 

If the belt speed is supposed to be equal to the speed of the pitch 
surfaces of the pulleys 

tN (D + 2 p) = rN 1 (A + 2 p), 

N_ = Di±2_ P 
0r Ni D + 2 P ' 



(16) 



That is, the angular speeds of the shafts are in the inverse ratio of the 
effective diameters of the pulleys, and this ratio is constant for circular 
pulleys. 



BELTS, ROPES AND CHAINS 



23 



As the thickness of belts generally is small as compared with the 
diameters of the pulleys, it may be neglected. 
The speed ratio will then become 

N_ Dx 

Ni D ' 

which is the equation almost always used in practical calculations. 

Example 4. Assume that a shaft A makes 360 r.p.m. On A is a pulley 24 ins. 
in diameter belted to a pulley 36 ins. in diameter on another shaft B. To find speed 
of shaft B. 



(17) 



From Eq. (17) 



Speed of A 
Speed of B 



Diam. of pulley on B 



Diam. of pulley on A 
Substituting the known values, this equation becomes 



360 



Therefore, 



Speed of B 24 
Speed of B = ff X 360 



240 r.p.m. 



Example 5. Suppose a shaft A making 210 r.p.m. is driven by a belt from a 30- 
in. pulley on another shaft B which makes 140 r.p.m. ; to find the size of the pulley 
on A. 

Using the principle of Eq. (17) 

Speed of A _ Diam. of pulley on B % 
Speed of B ~ Diam. of pulley on A 



Therefore, 



30 



= — or i 



30 X 140 
210 



20 ins. 



210 

140 x 
Then a 20-in. pulley is required on A. 

The relative directions in which the pulleys turn depend upon the 
manner in which the belt is put on the pulleys. The belt shown in Fig. 
25 is known as an open belt and 
the pulleys turn in the same 
direction, as suggested by the 
arrows. The belt shown in Fig. 
28 is known as a crossed belt 
and the pulleys turn in opposite 
directions as indicated. 

47. Kinds of Belts. The ma- 
terial most commonly used for 
flat belts is leather. For some kinds of work, however, belts woven from 
cotton or other similar material are used. When the belt is to be 
run in a place where there is much moisture, it may be made largely 
of rubber properly combined with fibrous material in order to give 
strength. 




24 ELEMENTS OF MECHANISM 

Leather belts are made by gluing or riveting together strips of leather 
cut lengthwise of the hide, near the animal's back. If single thick- 
nesses of the leather are fastened end to end, the belt is known as a 
single belt and it is usually about T \ in. thick. If two thicknesses 
of leather are glued together, flesh side to flesh side, the belt is known 
as a double belt and is from ■£$ to f in. thick. The manner of uniting 
the ends of the strips to form a belt, and of fastening together the 
ends of the belt to make a continuous band for running over pulleys, 
is very important. A detailed discussion of these features is not 
necessary, however, in the present study of the subject. 

Leather belts always should be run with the hair side against the 
pulleys, if possible. 

48. Power of Belting. The amount of power which a given belt 
can transmit depends upon its speed, its strength and its ability to 
adhere to the surface of the pulleys. The speed is usually assumed 
to be the same as the surface speed of the pulleys. The strength, of 
course, depends upon the width and thickness and upon the nature 
of the material of which the belt is made. The ability to cling to the 
pulley so as to run with little or no slipping depends upon the condi- 
tion of the pulley surfaces and of the surface of the belt which is in 
contact with the pulleys, and upon the tightness with which the belt 
is stretched over the pulleys. Since leather belts are more common 
and more nearly uniform in their character than those of other ma- 
terials, the discussion of power will be confined to them. 

49. Unit of Power — Horse Power. In order to measure the 
power, or the amount of work done, by any force, it is necessary to 
have some standard of measurement. A common unit for measur- 
ing work done is that known as the foot-pound. A foot-pound is the 
amount of work done in raising a one-pound weight a distance of 
one foot, or in moving any number of pounds through such a dis- 
tance that the product of the force exerted multiplied by the distance 
moved is equal to one. For example, if a 12-lb. weight is lifted one- 
twelfth of a foot, the work done is 12 lbs. X -^ ft. = 1 ft. -lb. 
If the apparatus furnishing the force to raise this weight is such that 
it can raise it in one minute, the apparatus is said to be capable of 
doing one foot-pound of work per minute, or to have a power of one 
foot-pound per minute. 

For measuring large quantities of power a larger unit is used, 
known as a horse power. One horse power is equal 33,000 ft. -lbs. of 
work per minute. For example, an engine which is capable of doing 
one horse power work is one which can move 1 lb. through a distance 
of 33,000 ft. per minute, or 33,000 lbs. 1 ft. per minute, or any num- 



BELTS, ROPES AND CHAINS 



25 




Fig. 29 



ber of pounds through such a distance in a minute that the product 
of the force multiplied by the distance moved in a minute is 33,000. 

50. Tension in a Belt. In Fig. 29 suppose the pulley A is fast to 
the shaft S and the pulley B fast to the shaft Si. Let it be assumed 
that when the shafts are at rest a belt is stretched over the pulleys 
as shown, the tightness with which it is stretched being such that 
there is a tension or pull in the belt of. a definite number of pounds. 
This tension is practically the 

same at all places in the belt 
and is called the initial tension. 
Let this initial tension be rep- 
resented by the letter T . 
Suppose now that some external 
force is applied to the shaft S 
causing it to tend to turn in 
the direction indicated by the arrow. This tendency to turn will 
increase the tension in the lower part of the belt (say between m and n) 
and decrease the tension in the upper part. Let the new tension in the 
lower or tight side of the belt be represented by Ti (which is greater 
than To) and the tension in the upper or slack side by T 2 (which is less 
than To). 

If the belt sticks to the pulley B so that there is no slipping, the 
force Ti tends to cause the pulley B to turn as shown by the full 
arrow and the force T 2 tends to cause B to turn as shown by the 
dotted arrow. As soon as T x becomes enough greater than T 2 to 
overcome whatever resistance the shaft Si offers to turning, the 
pulleys will begin to turn in the direction of the full arrow. The 
unbalanced force, then, which makes the driven pulley B turn is the 
difference between the tension Ti on the tight side of the belt and the 
tension T 2 on the slack side of the belt. This difference in tensions 
is called the effective pull of the belt and is here represented by the 
letter E. 

From the above discussion it may be seen that the following equa- 
tion holds true : 

T t -T 2 = E. (18) 

51. To Find the Horse Power of a Belt. Since, as explained in 
the previous paragraph, the effective pull is the force in the belt which 
enables it to do work, it follows that the product of the effective pull 
multiplied by the speed of the belt in feet per minute will give the 
foot-pounds of work per minute that the belt performs, and this divided 
by 33,000 will give the horse power which the belt transmits. If N is 



26 ELEMENTS OF MECHANISM 

the r.p.m. of S, and D the diameter of pulley A (in feet) the following 
equation expresses the horse power of the belt. 

Belt speed in ft. per minute X E _ TS p no . 

33^00 " aif ' Uyj 

irDN{T,-T 2 ) 

33,000 - H ' R (20) 

It is evident from the above that for a given belt speed the greater 
the difference between T x and T 2 the more horse power the belt trans- 
mits. Any figures which may be given for the maximum allowable 

T 

stress in a belt and the maximum ratio -^ are necessarily approximate 

and somewhat a matter of opinion. For the purpose of illustrat- 
ing the method of calculating the power which a given belt might 

be expected to transmit it will be assumed that ~- may not exceed % 

J 2 

and that the maximum allowable tension per inch of width is 140 

pounds for a double leather belt and 75 pounds for a single belt. If it 

is still further assumed that stresses due to centrifugal force may be 

neglected, 

Ti 7 

~r = 5 and T\ = 140 lb. per inch of width, 

1 2 o 

whence 

Ti-T 2 = 80 lb. 

= maximum effective pull per inch of width of double belt. 

Also T, = \ and Tl = 75, 

whence 

Ti - T 2 = 43 lb. (nearly) 

= maximum effective pull per inch of width of single belt. 

Substituting these values in Eq. (19), 
Belt speed in ft. per minute X 1$ X width of belt in inches 
38,000 

= H.P. a single belt will transmit (21) 
Belt speed in ft. per minute X 80 X width of belt in inches 
88,000 

= H.P. a double belt will transmit. (22) 

Corrections must be made in equations (21) and (22) if the belt speed 
is such that centrifugal force must be taken into account. It must also 
be borne in mind that the figures 48 and 80 are subject to modification. 



BELTS, ROPES AND CHAINS 27 

A simple and somewhat more conservative rule for estimating the 
power of a belt is known as the millwrights' rule and has been deter- 
mined largely by experience. This rule is as follows: 

A single belt traveling 1000 ft. per minute will transmit 1 H.P. per 
inch of width and a double belt traveling 560 ft. per minute will transmit 
1 H.P. per inch of width. 

Whence 

Belt speed in ft. per minute X width of belt in inches _ 
lOOO " ~ 

H.P. a single belt will transmit. (23) 
Belt speed in ft. per minute X width of belt in inches _ 
560 ~ 

H.P. a double belt will transmit. (24) 

Example 6. A shaft carrying a 48-in. pulley runs at a speed of 180 r.p.m. An 
8-in. double belt runs over the pulley and drives another shaft. To find the power 
that the belt can be expected to transmit without excessive strain. 

Solution 1. Using formula (22), 

Belt speed in feet = ^- X 180 = 2262 ft. per minute. 

„, 2262X80 X8 „ 01 „ _, , . . 
Then 33,000 = 43 " HR (nearly) ' 

Solution 2. Using formula (24) 

Belt speed as in solution 1 = 2262, 
2262 X : 



560 



= 32 H.P. 



It will be noticed that the two solutions given above give widely dif- 
ferent answers, that from the millwrights' rule being nearly 33 per cent 
less than the other. For the higher belt speeds this difference will not 
be as great if proper allowance is made for centrifugal force. Any 
such solution for a belt must be considered approximate, and merely 
furnishes a means of estimating the horse power roughly. There is no 
doubt that the above belt, if in proper condition, would carry much 
more than even the 43| H.P., but the heavier the belt is loaded the 
more attention it will require and the shorter will be its life. 

Example 7. A shaft running 200 r.p.m. is driven by a single belt on a 24-in. 
pulley. 15 H.P. is required. To find a suitable width of belt to use. 

Solution 1. Using formula (21), 

ir 24 
Belt speed = — -^ X 200 = 1257 ft. per minute. 

tw 12 57 X 43 X width 
Then ' 3P00 = 15 ' 

wj a u 15 X 33,000 _ . , 

Wldth= 1257X43 = 9in - neari y- 



28 ELEMENTS OF MECHANISM 



Solution 2. Using formula (23), 

1257 X width 
1000 



15, 



_,., 15,000 t _ . , 

Width = ' = 12 m. nearly. 

Here again the millwrights' rule shows a wider belt necessary for a 
given horse power. 

Example 8. Two shafts A and B are to be connected by a 12-in. double belt 
carrying 72 H.P. A is the driving shaft, making 240 r.p.m. B is to run 180 r.p.m. 
To find the size of the pulleys on A and B, using the millrights' rule. 

First find the necessary belt speed using Eq. (24). 

Belt speed X 12 _ 

560 ~ 7Z ' 

Belt speed = r^ = 3360 ft. per minute. 

Since B is to turn 180 r.p.m., if x = the diameter of the pulley on B, then, 
7T x X 180 = 3360 

or x = ^0- = 5.94 ft. = 71.28 in. 

180 7T 

or, since pulleys of that size would not be made in fractional inches, a 72-in. pulley 
would be used. 

Pulley on A _ 180 

Pulley on5~ 240 ' 

whence Pulley on A = 72 X |fg = 54 in. 

52. Approximate Formula for Calculating the Length of Belts. In 

finding the length of belt required for a known pair of pulleys at 
a known distance apart, the most satisfactory method, when possible, 
is to stretch a steel tape over the actual pulleys after they are in posi- 
tion, making a reasonable allowance (about 1 " in every 10 ft.) for stretch 
of the belt. Often, however, it is necessary to find the belt length from 
the drawings before the pulleys are in place or when, for some other rea- 
son, it is not convenient actually to measure the length. Various formu- 
las have been devised by which the length may be calculated when the 
pulley diameters and distance between centers of the shafts are known. 
These formulas, if exact, are all more or less complex and are, of course, 
different for crossed and for open belts. If the distance between 
shafts is large, the following will give an approximate value for the 
length of the belt. Referring to Fig. 31 and letting L represent the 
length of the belt, 

L = T(D + d ^ + 2 C. (25) 



BELTS, ROPES AND CHAINS 



29 



D, d and C, must be expressed in like linear units; if in feet, the result- 
ing value of L will be in feet; if in inches, the value of L will be in 
inches. 

In the case of an open belt where the two pulleys are of the same 
diameter the above formula gives an exact answer. If the pulleys are 
not of the same diameter, the length of belt obtained by Eq. (25) will 
be less than the correct length. If the shafts are several feet apart 
and the difference in diameters of the pulleys is not great, the percent- 
age error is very small for an open belt. With a crossed belt, pulleys 
of medium size and the shafts several feet apart, the result from the 
use of Eq. (25) is considerably less than the real length. This equation 
is accurate enough to use for an estimate of the length of a belt. 

53. Exact Formulas for Length of Belt Connecting Parallel Axes. 
While the methods given in the preceding paragraph are sufficient for 
the conditions there referred to, it is necessary in designing certain 
pulleys, known as stepped pulleys and cone pulleys, to make use of an 
equation expressing exactly, or very nearly so, the belt length in terms 
of the diameters and the distance between centers of the pulleys. The 
crossed belt and the open belt must be considered separately. 




Fig. 30 



Crossed Belts. Let D and d (Fig. 30) be the diameters of the 
connected pulleys; C the distance between their axes; L the length of 
the belt. 

Then 



where 



2 (mn + no + op) 




a 


-MJ.D + 2Ccos0 4 


(§■ 


(i 


+ d\(D + d)+2C 


3OS0, 


at 
ab 


an + bo D + d 

ab 2C 





(26) 



30 



ELEMENTS OF MECHANISM 



Open Belts. Using the same notation as for crossed belts, we 
have (Fig. 31) 



(27) 





L = 2 


(mn -\- no -\- op) 












= ($+ e ) D + 2Ccose + (%- 


-e\d 




= ?L(D + d)+d(D- 


-d) + 2C 


cos 6, 


where 


sin 6 = - 
cos 6 = \ 


n — bo D — d 
C 2C 






and 


1 (D-d)\ 




c 


^ f/77 a* 






Sp 




/ \ 




\ a t y f 




\B J 






\ i 


< G 




V0-* 


.b-^ 











Fig. 31 



For an open belt, 6 is generally small, so that 6 = sin 6, very nearly; 
then 



L = ^(D + 



+ (A=«! + 2CV / ) 



2(7 



(P ~ <*) 2 

4C 2 



nearly, 



= ^(Z) + d) + 2C 



0> - dy 

4C 2 



V 



1 ^-^—, nearly. 



If the quantity under the radical sign is expanded, and all terms hav- 
ing a higher power of C than the square in the denominator are neglected, 
since C is always large compared with (D — d), 



L=|(D+d)+2c{ ( -^ r ^+l 

L = |(D + d) + 2 C + (D 4 ~^ )2 > very nearly. 



(28) 



54. Stepped Pulleys. Sometimes it is necessary to have such a 
belt connection between two shafts that the speed of the driven shaft 
may be changed readily while the speed of the driving shaft remains 
constant. One method of accomplishing this is the use of a pair of 



BELTS, ROPES AND CHAINS 



31 



.*=. A 



pulleys each of which has several diameters as shown in Fig. 32. Such 
pulleys are known as stepped pulleys. Suppose that the shaft S, Fig. 
32, is the driver, making N r.p.m. When the belt is in the position 
shown in full lines, the working diameter of pulley A is Dx and the 
working diameter of pulley B is dx. 
Then if nx represents the r.p.m. of 
Si, when the belt is in this place, 

wi_ Di 

N dx' 

If the belt is shifted to any other 
position, as that shown by dotted 
lines, D x becomes the working diam- 
eter of the driving pulley and d x of 
the driven pulley. If n x represents 
the speed of S for this belt position 



D. 




Fig. 32 



Therefore, by properly proportioning 
the diameters of the different pairs 
of steps, it is possible to get any 
desired series of speeds for the driven 
shaft. 

In designing such a pair of pulleys 
two things must be taken into ac- 
count. First, the ratio of the diam- 
eters of the successive pairs of steps 
must be such as to give the desired 
speed ratios. Second, the sum of the diameters of any pair of steps 
must be such as to maintain the proper tightness of the belt for all 
positions. This second consideration makes the problem of design 
considerably more complicated. 

Two cases arise: First, the design of the pulleys for a crossed belt 
and second, the design for an open belt. 

55. Stepped Pulleys for Crossed Belt. Assuming that the value of 
D\, N, nx, n x and C are known for the drive shown in Fig. 32, and as- 
suming that the belt is crossed, instead of open as there shown, let it 
be required to find a method for calculating D x and d x . 

First find dx, which is readily done from the equation 
nx Dx 
N dx' 



32 



ELEMENTS OF MECHANISM 



in which dx is the only unknown quantity. Knowing, then, Dx and d L 
the value of Dx + dx is known. 
From Eq. (26) the length of the belt to go over the steps Dx and dx is 



(§ + *) 



(Di + dx) + 2 C cos e u 



When the belt is on the steps whose diameters are D x and d x the 
equation for the length of the belt is 



(^ + d x )(D x + d x ) + 2Ccos6 x . 



Since the same belt is to be used on both pairs of steps the value of 
these two equations must be the same. 

Therefore, 
(| + 0i) (Di + d x ) + 2Ccos^ = (|+0 (Z>* + 4) +2Ccos0*. 

Since C is a constant and 6 is dependent upon C and D + <i it follows 
that the above equation will be satisfied if 

D x + dx = Dx + dx . (29) 

Therefore in designing a pair of stepped pulleys for a crossed belt the 
sum of the diameters of all pairs of steps must be the same. 
Then from the equation 

Ux = Dx 

N d x 
and Eq. (29) 

D x + d x = Di + dx. 

D x and d x may be found by the method of simultaneous equations. 

Example 9. To find the diameters of all the steps in the pulleys shown in Fig. 
33 if a crossed belt is to be used. 



First, find d\ from the equation 



ni Dx 

N dx 



TTOTliqo 



whence 

Therefore, 
From Eq. (29) 

and 



192 = 16 

120 ~ di ' 

, 16X120 1A . 
* = -T92— =10m " 

Dx + -di = 16 + 10 = 26 in. JjJJ J>A 

D 2 + d 2 = A + dl = 26 in. ijJljjpll ggjjfcj; 



Dt = 160 
<k 120 



80 R.P.M.=n 3 
Fig. 33 



BELTS, ROPES AND CHAINS 33 

or A = f d 2 . 

Substituting this value of A in the preceding equation, 

I d 2 + d 2 = 26 
or ld» = 26, 

whence d 2 = — = — = 11? in. = 11.14 in. 

and A = 26 - 11+ = 14$ in. = 14.86 in. 

Again, A + d 3 = 26 in. 

A 80 
and Ik = 120 

or A = I d 3 , 

.-. |d 3 + d 3 = 26, 

or f (^3 = 26 in., 

whence d 3 = 15f in. = 15.6 in. 

and D 3 = 26 - 15| = 10| in. = 10.4 in. 

56. Stepped Pulleys for Open Belt. Referring still to Fig. 32, if the 
belt is open its length when on the steps Di and d t is, from equation (28), 

and when on steps D x and d x 

L = l(D z + d,)+2C+ (D '- c d - )2 - 

Equating these two expressions gives 

I CD, + 4) + ^*>! _ I (Du + <U + ( -^j=# • (30) 

This may be solved simultaneously with ^ = -~ to get the values of 

Da and d x * 

If the shafts are several feet apart and the range of speeds for the 
driven shaft is not excessive the diameters calculated for an open belt 
differ only very slightly from those for a crossed belt, and stepped 
pulleys designed for a crossed belt are often used for an open belt. If 
the shafts are close together and the speed range is large the crossed 
belt pulleys cannot be used for an open belt. 

* Equation (30) may be written in the form 

n . a n <a _i_ (A - dtf - (A - d x Y 
D x + d x = A + «i -\ ~9~C 

This may be solved approximately, in connection with ■— = -=— » by substituting for 

N d x 

(D x — d x ) 2 the value which it would have if the belt were crossed. 



34 



ELEMENTS OF MECHANISM 



Example 10. To find the diameters of all the steps in the pulleys shown in Fig. 
34, if an open belt is to be used. Shafts 24" on centers. 

First find d x from the equation 

rh = A 

N di 

900 18 

or — = — > 

150 di 

, 150 X 18 



whence 



900 



3 in. 



To find A and d 2 substitute in equation (30) the values of A, di and C, 



and 



n 2 A , 450 A r. o , 

whence v^ = — or A = 3 a 2 . 



^ d 2 — 150 d , 

Substituting this value for A and solving, 

d 2 = 5.43 in. A = 16.29 in. 

SimMy , | (18 + 3) + M - | (D , + « + M 



and 

Substituting and solving, 



n 3 A , 75 A , Q n 

at = T" ' whence r-^ = -y or d 3 = 2 A, 
TV d 3 150 d 3 



D 3 



in. and d 3 = 14.76 in. 



S" 



u 



ISO R,P.M,-U 



The proportion chosen in the data for Example 10 gives an extreme 
case, and it will be noticed that even here the amount that D 2 + di 

varies from Di + di is only about f in. 
and the variation of D 3 -+- d 3 is a trifle 
less than 1 T \ in. These quantities 
are large enough to affect the tightness 
of the belt and must, therefore, be 
taken into account. In ordinary cases, 
however, where the distance between 
centers is much larger than in Example 
10 and where the speed ratios are not 
so great the value of D x + d x , as 
obtained from Eq. (30) by the method 
just illustrated, differs but very little 
from Di + di and this difference can 
usually be neglected. 

57. Equal Stepped Pulleys. It is 
common practice, when convenient, 
to design a pair of stepped pulleys in such a way that both pulleys 
have the same dimensions and can, therefore, be cast from the same 




us, 

900 R. P.M. =77, 

4-50 H.P.M.=*a 2 
75R,P:U.-a 3 

34 



BELTS, ROPES AND CHAINS 



35 



pattern. This condition imposes certain restrictions on the speed ratios 
as may be seen from the following: 



Referring to Fig. 35 if the pulleys are alike, 

Di = d 5 , D 2 = d 4 , D 3 = d 3 , Z> 4 = d 2 , D 5 
As in previous discussions, 



4. 

















N di 


and 


n 5 D h 
N d 5 


but 


D, dt 
d 5 Dt 


Therefore, 


7h N 

N ng" 


In a similar manner, 


?h_N 
N ni 


and 


N = n 3 . 




D, 


D 2 


D 3 


D 4 


D 5 


VI 
















N-R.P.M, 



















(3D 



Q, J M.z,ja <u j3 }KP n ! rR. P. §. 
Fig. 35 



A 


D 2 


A. 


</i 


d 2 


d, 


") 







n 3 =60 



Fig. 36 



That is: When equal stepped pulleys are used the speeds of the driven shaft 
must be so chosen that the speed of the driving shaft is a mean proportional 
between the speeds of the driven shaft for belt positions symmetrically 
either side of the middle step. 

Example 11. A pair of equal three-stepped pulleys, Fig. 36, are to carry a 
belt to connect two shafts. The driving shaft makes 120 r.p.m., and the lowest 



36 



ELEMENTS OF MECHANISM 



speed of the driven shaft is 60 r.p.m. To find the other two speeds of the driven 

shaft. 

n_i _ N 
N~ n z 

«L = 120. 
or 120 60 ' 

Therefore «i = 240, 

ri2 = N = 120. 

If the step diameters are to be calculated, it will be done by the methods explained 
in § 55 or § 56 according as the belt is crossed or open. 




Fig. 37 



58. Speed Cones. In some cases instead of stepped pulleys, 
pulleys which are approximately frusta of cones are used, as shown in 
Fig. 37. Here the working diameters of the pulleys, as D x and d x for 
any belt position, are measured at the middle of the belt. To design 



BELTS, ROPES AND CHAINS 



37 



such a pair of pulleys a series of diameters D h D 2 , D 3 , etc. (Fig. 38), 
may be calculated in the same way as steps and plotted at equal dis- 
tances (a) apart, then a smooth line drawn through their ends, as 
shown. The length (a) does not affect the problem except as it makes 
the cone longer or shorter. The contours may be straight lines as in 
Fig. 38, giving cones, or curves as in Fig. 39, giving conoids. 

When cones are used, a shipper must guide each part of the belt just 
at the point where it runs on to the pulley (see Fig. 37) ; otherwise 
the belt will tend to climb toward the large end of each pulley. 
Both shippers must be moved simultaneously when the belt is 
shifted. 




cS Q cS"' 


\ 





Fig. 38 



Fig. 39 



59. Belt Connections between Shafts which are not Parallel. 

Non-parallel shafts may be connected by a flat belt with satisfactory 
results, provided the pulleys are so located as to conform to a funda- 
mental principle which governs the running of all belts, namely: The 
point where the pitch line of the belt leaves a pulley must he in a plane 
passing through the center of the pulley toward which the belt runs. 
In other words, a plane through the center of the receiving pulley, per- 
pendicular to the pulley axis, must, if produced, include the delivering 
point of the pulley from which the belt is running. This may be seen 
by a reference to Fig. 40. In this case the shafts S and T are intended 
to turn in the directions indicated by the arrows. Considering Eleva- 
tion A, the pitch line of the belt leaves the pulley M at the point a. If 
the pulley N is in such a position on the shaft T that a plane through 
the middle of its face contains the point a, the belt will run properly 
on to pulley N. XX is the trace of this plane and evidently contains 
point a. Similarly, in Elevation B, the pitch line of the belt leaves the 
pulley N at 6i and M is so located on shaft S that a plane YY through 
the middle of its face contains 6i. 

Fig. 41 shows the proper relative position of the pulleys if the direc- 
tion of turning of shaft S is the reverse of that in Fig. 40. Other 
changes in the directions of rotation of either pulley would necessitate 
corresponding changes in the relative positions. 



38 



ELEMENTS OF MECHANISM 



Both Figs. 40 and 41 show the pulleys at 90° with each other. The 
belt would run equally well if the pulleys were turned at any angle about 
XX as an axis. 




Elevation A 



Elevation B 



Fig. 40 



60. Quarter-turn Belt. A belt which connects two non-intersect- 
ing shafts at right angles with each other, similar to those in Figs. 40 
and 41, is called a quarter-turn belt. Emphasis should be laid on the 
fact that, for any given setting of the pulleys, the shafts must always 



BELTS, ROPES AND CHAINS 



39 



turn in the direction in which they were designed to turn. If the 
direction of rotation is changed without resetting the pulleys, the belt 
will immediately leave the pulleys. For this reason simple quarter- 




Fig. 41 



turn belts like those illustrated above are likely to give trouble if 
used in places where there is possibility of the shafting turning back- 
wards even a small fraction of a turn. If this should happen to a 



40 



ELEMENTS OF MECHANISM 



small belt, it could easily be replaced on the pulleys; in the case of a 
large belt, however, the replacing would be more difficult. 

61. Reversible Direction Belt Connection between Non-parallel 
Shafts — Guide Pulleys. If the connection between two non- 
parallel shafts is to be such that the shafting may run in either direc- 
tion and still have the pulleys deliver the belt properly, in accordance 
with the fundamental law already explained, it is necessary to make 
use of intermediate pulleys to guide the belt into the proper plane. 
Such pulleys are called guide pulleys. 

62. Examples of Belt Drives — Method of Laying Out. The fol- 
lowing examples will illustrate a few of the types of belt drives which 

may occur and will give some 



jSi 



f* 



-G* 



Zl 



Front Elevati 



Right Elevation 



Fig. 42 



idea of the method of pro- 
cedure in designing such drives. 
Some of these examples are 
chosen from existing drives; 
others have been modified in 
order to illustrate the principles 
more clearly. 

Example 12. Given two shafts A 
and B located as shown in Fig. 42. 
Shaft A, carrying a 52-in. pulley, is 
to drive a 60-in. pulley on shaft B by 
means of a 12-in. double belt. Two 
guide pulleys 30 ins. diameter in line 
with each other are to be located on 
two horizontal shafts so that the 
direction of rotation may be reversed 
without the belt running off. When turning in the direction indicated by the 
arrows, the tight side of the belt is to run direct from the driven to the driving pulley 
in a vertical line, the loose side returning around the guide pulleys. The guide pulley 
which receives the belt from the upper main pulley is to be on a shaft whose center 
is 9 ft. below the center of A ; the other guide pulley is to be on a shaft whose center 
is 2 ft. 6 in. below the center of A. Main pulleys 16-in. face, guide pulleys will be 
drawn the same width as the belt although in practice they would be wider. 
To draw two elevations and a plan. 

Solution. Referring to Fig. 43, first draw the center line YYi which is the center 
line through the shaft B. At a distance of 14 ft. above YYi draw XX-i. as the center 
line of shaft A. At any convenient place near the left end of YYi choose the point 
Bl as the center point of shaft B for the left elevation. With Bl as a center draw 
a circle 60-ins. diameter which will be the left elevation of the pulley on B. Next, 
draw the vertical line TT\ tangent to this pulley on the side which is moving up- 
ward. TT X is the pitch line of the tight part of the belt and must be contained in 
the center plane of the pulley on A . With TTi and XXi as center lines, draw the 
rectangle 1-2-3-4 of length equal to the diameter of the pulley on A and width 
equal to the width of face of the same pulley. This rectangle is the side view or 



BELTS, ROPES AND CHAINS 



41 



AjtE 



fE- 




Fig. 43 



42 



ELEMENTS OF MECHANISM 



left elevation of the upper or driving pulley. Next, choose a point Ap near the right 
end of XXi and draw the end view or right elevation of this pulley. The vertical 
line Tip Tp drawn tangent to the upward moving or driving side of this pulley will 
be the right elevation of the driving or tight part of the belt and must be contained 
in the center plane of the driven pulley, since the direction of rotation is to be re- 
versible. Therefore, the lines T x p T P and YY\ are the center lines for the rectangle 
5-6-7-8 which is the right elevation of the driven pulley, TT X and T P T lP are the 
two views of the line of intersection of the center planes of the pulleys. 

The position of the two main pulleys with respect to each other has now been 
determined and the two elevations drawn. Their plan must next be drawn. This 
may be placed above either elevation, and is here placed above the left elevation. 
Looking down on the pulleys, both will appear as rectangles. The center line 
XnXih may be drawn at any convenient distance above XXi and is the horizontal 
projection, or plan view, of the center line of the shaft A. The pulley on this center 
line can be projected directly up from the rectangle l-2-3-4^and will, of course, have 
the same dimensions. The center line MN of the shaft B will be vertically above 
Bl and the rectangle which forms the plan view of the pulley on B will be located 
on this center line with its middle line passing through the front end of the plan of 
the other pulley. In other words, the plan view of these two pulleys is obtained by 
projecting from the two elevations in accordance with the ordinary principles of 
projections. 

To draw the guide pulleys, first locate them in the plan. Their center plane will 
contain the line VW and one will have its contour passing through point V in plan 
while the contour of the other will pass through 
W. To draw the elevations, first draw the center 
lines at the specified distances below shaft A and 
then draw the ellipses which represent the pulleys 
by projecting from the plan. 

Example 13. Shaft S (Fig. 44) drives shaft 
T by means of an 8-in. double belt. Both main 
pulleys 36 in. diameter located as shown. The 
usual direction of rotation to be as indicated by 
the arrows but the arrangement to be such that 
the directions may be reversed. Two 15-in. 
guide pulleys are to be placed on a vertical shaft 
to carry the belt between the two main pulleys. 
All pulleys 9 ins. face. 

To draw the drive, making the elevations 
and a plan. 

Solution. (See Fig. 45.) Draw the three views of the main shafts and pulleys, 
the plan and front elevation being the same as shown in Fig. 44, and the right eleva- 
tion being constructed from these in accordance with the usual principles of projec- 
tion. The left elevation might have been made instead of the right. 

The position of the guide pulley shaft can best be determined from the plan. 
The planes of the pulleys A and B intersect in a line which, in plan, is projected as 
the point P and in the two elevations as the lines XY and XiYi, respectively. Since 
the upper guide pulley is to deliver the belt to pulley B, it must be tangent to the 
line PM, and since, if the direction of rotation is reversed, C must be able to deliver 
the belt to pulley A it must be tangent to the line PN. The same reasoning will 
apply to the lower guide pulley. The center of the guide pulley shaft will, there- 




Front Elevation 
Fig. 44 



BELTS, ROPES AND CHAINS 



43 



fore, be at a point which is distant from PM and PN an amount equal to the radius 
of the guide pulleys. Since the pulleys A and B are of the same diameter and their 
axes on the same level, the guide pulley C will appear in the elevations with its 
center plane tangent to the tops of A and B, and D will have its center plane tan- 
gent to the bottoms of A and B. With this arrangement it is possible for either of 




Front Elevation 



Right Elevation 



Fig. 45 



the main pulleys to deliver the belt into the plane of either guide pulley, and either 
guide pulley may deliver to either main pulley. 

It should be noticed that a drive like this, with both guides on the same vertical 
shaft, can be reversible in direction only when the main pulleys are of the same 
diameter. The next two examples show the construction when the main pulleys 
are of different diameters. 

Example 14. Referring again to Fig. 44, assume the same conditions as for 
Example 13, except that the main pulleys are of different diameters. Suppose the 
pulley on T is 48-in. diameter and that on S 36-ins. diameter. The direction of rota- 
tion not to be capable of being reversed. 

Solution. See Fig. 46. The three views of the main shafts and pulleys are drawn 
as in Example 13. The center of the guide pulley shaft is located in the plan at 
such a point that the pulley circumference will be tangent to the lines PM and PN 
as in Fig. 45. The position of the guide pulley C on this shaft is determined in the 
front elevation, it being at such a height that its center plane will be tangent to the 



44 



ELEMENTS OF MECHANISM 



top surface (that is, contain the point of delivery K) of the pulley A which delivers 
the belt to C. Similarly, in the right elevation, the position of the guide pulley D 
is such that its center plane will be tangent to the lower surface (that is, will contain 
the point of delivery R) of the pulley B. 




Front Elevation " 

Fig. 46 



Yi Right Elevat'n 



Example 15. With the data the same as for Example 14, suppose it is required 
so to arrange the guide pulleys that the direction of rotation may be reversed. (They 
cannot in this case be on the same vertical shaft.) 

Solution. See Fig. 47. After having drawn the three views of the main shafts 
and pulleys, the problem becomes one of so placing the guide pulleys that they will 
conduct the belt in either direction. There are a great many possible solutions of 
this problem, but that shown in Fig. 47 is the simplest. 

In the front elevation the points a and b are the center points of the upper and 
lower contour elements of the pulley B. From a and b draw lines ae and bf tangent 
to pulley A. The center planes of the guide pulley C must contain the line ae and 
the center plane of the guide pulley D must contain the line bf. C will appear in 
this view, therefore, as a rectangle with one end passing through a and D will appear 
as a rectangle with one end passing through b. In the other views the edges of the 
guide pulleys will appear as ellipses, as shown. 

Example 16. The shaft S, Fig. 48, is to drive the shaft T by means of an 8-in. 
belt running on pulleys A and B. One of the columns of the building makes it im- 
possible to continue the shaft T far enough to place the pulley B in the proper posi- 



BELTS, ROPES AND CHAINS 



45 



tion relative to A to permit the use of a direct quarter-turn drive. Furthermore the 
vertical distance between the two shafts is too small to make such a drive practicable 
even if the column did not interfere. It is, therefore, necessary to employ guide 
pulleys to conduct the belt from B to A and from A back to B. The relative direc- 




FRONT ELEVATION 



Y 1 RIGHT ELEVATION 



Fig. 47 



tions of rotation are to be as shown and the guide pulleys are to be so located that the 
directions may be reversed. 18-in. guide pulleys will be used. 

Solution. See Fig. 49. The positions of the guide pulleys are determined from 
the left elevation. From the center points a and b of the upper and lower contour 
lines of pulley A, lines ae and bf are drawn tangent to the pulley B. The center 
plane of one guide pulley C must contain the line ae and the pulley will appear in 
this view as a rectangle with one end passing through a. The center plane of the 
other guide pulley D must contain the line bf and the pulley will appear as a rectangle 
with one end passing through b. The guide pulleys will appear in the front elevation 
and in the plan with their edges ellipses, as shown. The two guide pulleys so nearly 
coincide in the plan that the lower one was omitted in the drawing. 

It will be noticed that in all the preceding examples the same surface of the belt 
comes in contact Avith the main pulleys at all times. This is an important condi- 
tion from a practical standpoint. Whenever practicable the same surface should 
run against the guide pulleys also. 

Example 17. Given two shafts at right angles,, located as shown in Fig. 50. 
Shaft A carries a 52-in. pulley which drives a 60-in. pulley on shaft B by means of 



46 



ELEMENTS OF MECHANISM 









1 



T T 
Left Elevation Front Elevation 

Fig. 48 




FRONT ELEVATION 



& 



& 



B £ 

X 



BELTS, ROPES AND CHAINS 47 

a double belt 12 ins. wide. The ordinary direction of rotation is as shown by the 

arrows. One guide pulley 30 ins. diameter is to be so located that the direction of 

rotation may be reversed without the 

belt running off. When turning in the _g 

direction shown, the tight side of the 

belt is to run direct from driven to 

driving pulley in a vertical line, the 

loose side returning around the guide 

pulley. 

The main pulleys are 14 ins. wide. 
Two elevations and a plan are to be 
drawn. 

Solution. It will be noticed that, 
except for the guide pulley, this problem 
is the same as Example 12 and the 
method of drawing the three views of 
the main pulleys is exactly as described 
for that case 

To draw the guide pulley proceed as F r <i0 

follows: (See Fig. 51). The distance of 

this guide from either one of the main pulleys would be governed somewhat by con- 
venience in actually setting up the bearings to support it, and partly also by the 
relative sizes of the main pulleys. It is desirable so to locate it as to give the least 
possible abruptness to the bend in the belt. In this case there has been selected a 
point C in the line of intersection of the two main pulley planes which is 6 ft. 6 ins. 
below the axis of the upper shaft. This point will be at Cl and Cp in the two eleva- 
tions. From Cl draw a line tangent to the lower pulley at Dl and project across, 
getting the other view of this point at D P . In a similar way draw a line from Cp 
tangent to the upper pulley at E P and project across to find El. We now have the 
two projections of two hnes CD and CE drawn from a point in the intersection of 
the pulley planes tangent to the two pulleys, and the guide pulley must be set in 
such a position that its center plane will contain these two hnes. The problem then 
is to draw the projections of the guide pulley when so set. Either elevation may 
be drawn first; let us start with the left elevation. Here ClDl shows in its true 
length and the line CD itself may be considered as lying in the plane of the paper. 
The line CE has one end Cl in the plane of the paper while the line itself really 
slants down below the paper. The true size of the angle which it makes with the 
plane of the paper is equal to the angle EpCpFp. DlCl is the trace, or line of in- 
tersection with the paper, of the plane containing CD and CE. 

Now swing this plane up into the paper to get the true angle between CD and 
CE. To do this produce DlCl to the left and through El draw a line perpendicular 
to DlCl. From Cl with a radius equal to the true length of CE (that is, with radius 
CpEp) cut this perpendicular at E x and join Ei to Cl. The angle EiClD l is the true 
angle between the lines EC and DC. Set the compasses with a radius equal to the 
radius of the guide pulley and find by trial the center Oi about which a circle of 
this radius may be drawn tangent to ClEi and ClDl. This point shows the real 
position of the center of the middle circle of the guide pulley relative to the lines 
CE and CD. The next step is to revolve Oi back to find its projection relative to 
ClE l and ClDl. To do this draw a line through Ei and Oi meeting ClDl at Hi; 
then join Hi to El. Through Oi draw a line perpendicular to DlCl meeting HiEl 
at Ol and ClDl at Jl. Then Ol is the projection of the center point of the pulley. 



48 



ELEMENTS OF MECHANISM 
/fa 







Fig. 51 



BELTS, ROPES AND CHAINS 



49 




To draw the projections of the edges of the pulley select any point as </ 2 in the line 

ClDl, draw a line through Ol parallel to ClDl and from J 2 with a radius equal to 

the radius of the guide pulley cut this line at 2 . Draw a line through J 2 and O2, 

also a line V2W2 through 2 perpendicular to J 2 2 . On these two lines as center 

lines construct the rectangle TO 2 n2r2S2 of length equal to the diameter of the guide 

pulley and width equal to the width of face of the same. This rectangle is the side 

view of the guide pulley when revolved up into 

the plane of the paper, the line V2W2 being the 

center line of its shaft. The projection of the 

pulley consists of two equal ellipses with major 

axes equal to the diameter of the guide pulley 

and minor axes found by projecting the points 

m 2 n 2 and s 2 r 2 on to the line OlJl, as shown. If a 

definite length V2W2 is chosen for the shaft, the 

projections of the ends of the shaft are found at 

Vl and Wl. The right elevation of the guide 

pulley and the axis of its shaft are found by a 

method exactly similar to that just described. 

The plan view of the guide pulley is found by 
first finding the plan projections Vn and Wh of 
the two ends of the shaft by projecting up from 
the two elevations, then revolving this line VhWn 
over until it comes into the plane of the paper, 
as shown at FsWV On this line is drawn the 
rectangle t s x 3 y s Z3, representing the pulley, and the 
axes of the ellipses which constitute the plan view 
of the pulley are found by projecting the rectan- 
gle Uxzy%zs on to VnWh, as shown. 

63. Crowning of Pulleys. If a belt 
is led upon a revolving conical pulley, it 
will tend to lie flat upon the conical 
surface, and, on account of its lateral 
stiffness, will assume the position shown 
in Fig. 52. If the belt travels in the direction of the arrow, the point a 
will, on account of the pull on the belt, tend to adhere to the cone and 
will be carried to b, a point nearer the base of the cone than that pre- 
viously occupied by the edge of the belt : the belt would then occupy the 
position shown by the dotted lines. Now if a pulley is made up of two 
equal cones placed base to base, the belt will tend to climb both, and 
would thus run with its center line on the ridge formed by the union of 
the two cones. In practice pulley rims are made slightly crowning, 
except in cases where the belt must occupy different parts of the same 
pulley.* In Fig. 52 two common forms of rim sections are shown at C 
and D; that shown at C is most commonly met with, as it is the easier 

* The amount of crowning varies from about ■&" on a pulley 6" wide to about £" 
on a pulley 30" wide. 



f 



J 



Fig. 52 



50 



ELEMENTS OF MECHANISM 



to construct. When pulleys are located on shafts which are slightly 
out of parallel, the belt will generally work toward the edges of the 
pulleys which are nearer together. The reason for this may be seen 
from Fig. 53. The pitch line of the belt leaves pulley A at point a. 
In order to contain this point the center plane of pulley B would have 
to coincide with XXi. That is, the belt is delivered from A into the 
plane XX X . Similarly, the belt is delivered from b, on the under side 
of pulley B, into the plane Y{Y. The result of this action is that 
the belt works toward the left and tends to leave the pulleys. 

64. Tight and Loose Pulleys are used for throwing machinery into 
and out of gear. They consist of two pulleys placed side by side upon 
the driven shaft CD (Fig. 54); A, the tight pulley, is keyed to the 
shaft; while B, the loose pulley, turns loose upon the shaft and is kept 
in place by the hub of the tight pulley and a collar. The driving shaft 
carries a pulley G, whose width is the same as that of A and B put 
together, or twice that of A. The belt, when in motion, can be 
moved by means of a shipper that guides its advancing side, either 
on to the tight or the loose pulley. The 
pulley G (Fig. 54) has a flat face, because 
the belt must occupy different positions 
upon it, while A and B have crowning 
faces, which will allow the shifting of the 
belt and will retain it in position when 
shifted upon them. 

rh 





Fig. 53 Fig. 54 

65. Ropes and Cords. Power is often transmitted by means 
of ropes running over pulleys, called sheaves, having grooved 
surfaces. For large amounts of power inside of buildings the 
ropes are made of hemp or similar material. For long dist- 
ance drives and drives which are exposed to the weather wire 



BELTS, ROPES AND CHAINS 



51 



ropes are used. For small amounts of power on machines, cords of 
cotton are common. 

66. Systems of Driving with Hemp Rope. There are two distinct 
systems of rope driving, each of which has its advantages. One is 




Fig. 55 



the Multiple Rope or English System. This is the simpler of the 
two and consists of independent ropes running side by side in grooves 
on the pulleys. A large drive using this system is shown in Fig. 55. 



Guide Sheaves 




Fig. 56 

The other system is the Continuous or American System, shown 
in Figs. 56 and 57. One rope is wound around the driving and driven 
pulleys several times, and conducted back from the last groove of one 



52 



ELEMENTS OF MECHANISM 

■Driving 



Driven\ 




Loose Sheave' 



Fig. 57 




Fig. 58 



Fig. 59 



Fig. 60 



BELTS, ROPES AND CHAINS 



53 



pulley to the first groove of the other pulley by means of one or more 
intermediate pulleys which also serve the purpose of maintaining a 
uniform tension throughout the entire rope. The slack should be taken 
up on the loose side just off the driving sheave. There are two ways 





Fig. 61 



Fig. 62 



of accomplishing this. First (see Fig. 56), the rope is conducted from 
an outside groove of the driver to the tension sheave and after passing 
around it is returned to the opposite outside groove of the driven sheave. 
Second (see Fig. 57), where it is inconvenient to take the slack directly 
from the driver the rope is passed around a loose sheave on the driven 
shaft, thence over the tension sheave, and is returned 
to the first groove in the driven sheave. Figs. 58, 59 
and 60 show further examples. 

67. Grooves for Hemp Rope. The shape and 
proportions of the grooves used on many pulleys for 
hemp rope depend somewhat upon the system used. 
Figs. 61 and 62 show two forms much used. Fig. 63 
illustrates the groove used on idle wheels. 

It will be noticed that the rope wedges into the 
grooves on the driving and driven pulleys, while on the 
loose or idle pulleys it rides on the bottom of the groove. 

68. Small Cords are often used to connect non- 
parallel axes, and very often the directional relation 
of these axes must vary. The most common example 

is found in spinning frames 
and mules, where the spin- 
dles are driven by cords 
from a long, cylindrical 
drum, whose axis is at right 
angles to the axes of the 
spindles. In such cases, the common perpendicular to the two axes must 
be contained in the planes of the connected pulleys; both pulleys may 
be grooved, or one may be cylindrical, as in the example given above. 
Fig. 64 shows two grooved pulleys, whose axes are at right angles to 
each other, connected by a cord which can run in either direction, pro- 




Fig. 63 




54 



ELEMENTS OF MECHANISM 



vided the groove is deep enough. To determine whether a groove has 
sufficient depth in any case, the following construction (Fig. 65) may be 
used. Let A B and A\B X be the projections of the approaching side of 
the cord; pass a plane through A B parallel to the axis of the pulley; it 
will cut the hyperbola CBD from the cone FEG, which forms one side of 
the groove. The cord will lie upon the pulley from B to I, where it will 
leave the hyperbola on a tangent. If the tangent at / falls well within 
the edge of the pulley at C, the groove is deep enough. It will usually 
be sufficient to draw a straight line, as ab (Fig. 64), and see that it falls 
well inside of the point corresponding to C in Fig. 65. 




Fig. 65 



69. Drum or Barrel. When a cord does not merely pass over a 
pulley, but is made fast to it at one end and wound upon it, the 
pulley usually becomes what is called a drum or barrel. A drum for 
a round rope is cylindrical and the rope is wound upon it in helical coils. 
Each layer of coils increases the effective radius of the drum by an 
amount equal to the diameter of the rope. A drum for a flat rope 
has a breadth equal to that of the rope, which is wound upon itself in 
single coils, each of which increases the effective radius by an amount 
equal to the thickness of the rope. 

70. Wire Ropes. Wire rope is well adapted for the transmission 
of large powers to great distances, as for instance in cable and in- 
clined railways. Its rigidness, great weight, and rapid destruction 
due to bending, however, unfit it for use in mill service, where the 
average speed of rope is about 4000 ft. per minute. As the easiest 
way to break wire is by bending it, ropes made of it, by any method 



BELTS, ROPES AND CHAINS 



55 



whatsoever, have proved unsatisfactory for drives of short centers and 
high speed unless the diameters of the sheaves are large enough to avoid 
bending the rope to strain it above the elastic limit. 

Wire ropes will not support without injury the lateral crushing due 
to the V-shaped grooves; hence it is necessary to construct the pulleys 
with grooves so wide that the rope rests on the rounded bottom of the 
groove, as shown in Fig. 66, which shows a section of the rim of a wire- 
rope pulley. The friction is greatly increased, and the wear of the 




rope diminished, by lining the bottom of the groove with some elastic 
material, as gutta-percha, wood or leather made up in short sections 
and forced into the bottom of the groove. 

71. Chains are frequently used as connectors between parallel 
axes and also for conveying and hoisting machinery and for other 
similar purposes. The wheels over which chains run are called sprockets 
and have their surfaces shaped to conform to the type of chain used. 

Chains may be classified as follows: 

1° Hoisting chains 

2° Conveyor chains 



Detachable or Hook Joint 
Closed Joint 



f Block 

3° Power transmission chains I Roller 

I Silent 

72. Hoisting Chains. The most common form of hoisting chain 
consists of solid oval links as shown in Fig. 67. The form of sprocket 
used for such a chain is evident from the figure. 

73. Conveyor Chains may be of the detachable or hook joint type 
as shown in Fig. 68, or of the closed joint type illustrated in Fig. 69. 



56 



ELEMENTS OF MECHANISM 




Fig. 67 




Fig. 68 



Squared 



y=M= 



I 



^=^ 



Fig. 69 




BELTS, ROPES AND CHAINS 



57 



The design of the sprocket teeth is largely empirical, care being 
taken to have the teeth so shaped and spaced that the chain will run 
on to and off from the sprockets smoothly and without interference 
even after it has stretched or worn somewhat. Chains of this general 
class are often used for transmitting power at low speeds, as in agricul- 
tural machinery. They are usually made of malleable cast links and 




do not have the smooth running qualities of the more carefully made 
chains. 

74. Power Transmission Chains. This class includes the three 
types known as block, roller and silent. The chains are made of 
steel, accurately machined, with wearing parts hardened, and run on 
carefully designed sprockets. In the following discussion no attempt 
is made to give an exhaustive treatment of the subject, but merely 
to give some idea of the character of the three types and some of the 
points which need to be considered in their design. 

75. Block Chains. Fig. 70 shows a block chain made by the 
Diamond Chain & Mfg. Co. 



58 



ELEMENTS OF MECHANISM 

I 




Fig. 73 

Chains of the block type are less expensive to make than the roller 
or silent chains and are used for the transmission of power at com- 
paratively low speeds. They are also used to some extent as conveyor 
chains and for other purposes in place of the malleable chains of class 2. 



BELTS, ROPES AND CHAINS 



59 



Fig. 71 shows a block chain in place on the driving sprocket. Atten- 
tion is called to the way in which the links swing into position as they 
approach the sprocket and swing out as they leave. A method of 
laying out the sprocket teeth is indicated on the same figure. The 
proportions for the teeth here shown are those recommended by Mr. 
B. D. Pinkney in " Machinery/' January, 1916. 

76. Roller Chains. Fig. 72 illustrates a form of roller chain similar 
to one made by the Diamond Co., and Fig. 73 shows the same chain 





H — 




3»H 

/ 


^V 3 1 X 




^a 


U?_ — ^D— 


\i 




*o 



o 



o 



^7 



Fig. 74 



in place on the sprocket. The method of laying out the sprocket teeth 
in Fig. 73 is the one recommended by the Diamond Co., for roller chain 
sprockets. 

77. Calculations for Chain Length. This paragraph and the two 
following with Figs. 74, 75 and 76 are taken directly from "Power 
Chains and Sprockets" published by the Diamond Chain & Mfg. Co. 

D = Distance between centers. 

A = Distance between limit of contact. 

R = Pitch radius of large sprocket. 

r = Pitch radius of small sprocket. 
N = Number of teeth on large sprocket. 

n = Number of teeth on small sprocket. 
P = Pitch of chain and sprocket. 
(180° + 2 a) = Angle of contact on large sprocket. 
(180° — 2 a) = Angle of contact on small sprocket. 
. t R-r 



a 



sin" 



A = D cos a. 



60 



ELEMENTS OF MECHANISM 



Total length of chain. 

T _ 180 + 2 a 
L ~ 360 



NP + 



180 - 2 a 
360 



nP + 2 D cos a. 



(32) 



78. Calculations for Diameters of Sprockets for Block Chains. 

N = Number of teeth. 

b = Diameter of round part of chain block (usually 0.325). 
B = Center to center of holes in chain block (usually 0.4). 
A = Center to center of holes in side links (usually 0.6). 

180° 

a= -w 



jl an ju 


-7 + cos a 




Pitch diameter 


A 

sin B 


(33) 


Outside diameter 


= Pitch diameter + b. 


(34) 


Bottom diameter 


= Pitch diameter — b. 


(35) 



In calculating the diameter of sprocket wheels, the bottom diameter 
is the most important. 




Fig. 75 



Fig. 76 



79. Calculations for Diameters of Sprockets for Roller Chains. 

Referring to Fig. 76, 



BELTS, ROPES AND CHAINS 



61 



a = 



Pitch diameter 



(36) 



N = Number of teeth in sprocket. 
P = Pitch of chain. 
D = Diameter of roller. 
180° 
N 
P 
sin a 

Outside diameter = Pitch -f- D. (37) 

Bottom diameter = Pitch — D. (38) 

80. Silent Chains. None of the above mentioned chains can be 
run at high speed without noise. There are now in use several makes 
of chains known as Silent Chains which run satisfactorily at high 
speeds and which adapt themselves to the sprocket after the pitch 
of the chain has increased due to wear. Two examples will serve to 
illustrate this type of chain. 

81. Renold Silent Chain. Fig. 77 shows a chain developed by 
Hans Renold. It consists of links C of a peculiar form with straight 




bearing edges a, b, which run over cut sprocket-wheels with straight- 
sided teeth whose angles vary with the diameter of the wheel. The 
chain may be made any convenient width, the pins binding the whole 
together. One sprocket of each pair is supplied with flanges to retain 
the chain in place. The upper drawing shows a new chain in position 
on its sprocket, the bearing parts of the links being on the straight 
edges of the links only, not on the tops or roots of the teeth. The 
chain thus adjusts itself to the sprocket at a diameter corresponding 
with its pitch, and as any tooth comes into or out of gear there is neither 



62 



ELEMENTS OF MECHANISM 



slipping nor noise. The lower figure shows the position taken by a 
worn chain of increased pitch on the same wheel. 

82. Morse Rocker- Joint Chain. This chain (Fig. 78) eliminates 
the sliding friction of the rivets as the chain bends around the sprocket. 
Instead of the ordinary pin bearing a rocking bearing is provided at 
each joint. The following description, with slight changes, is taken 
from the catalogue of the Morse Chain Co. Two pins are employed 
at each joint; the left hand pin a is called the seat pin and the right 
hand pin b the rocker. Each is securely held in its respective end of 
the link. The seat pin has a plane surface against which the edge of 
the rocker pin rocks or rolls when the chain goes on and off the sprockets. 
The joint is so designed that the pressure due to tension of driving will 
be taken on a flat surface when in between the sprockets. 




Fig. 78 



Fig. 78 shows the chain on a driving sprocket running in the direc- 
tion indicated by the arrows. The angle of the tooth to the line of 
the pull and any centrifugal force that may exist both tend to keep 
the link out to its true pitch diameter during the revolution of the 
wheel ; it will fall below this point only when the pull of the slack side 
of the chain is greater than the forces in the opposite direction. 

From this it will be seen that there are two forces definitely opera- 
tive to keep the chain in its proper pitch contact with the wheels by 
causing it to assume a larger and larger circle as the chain lengthens in 
pitch; thus, the driving load continues to be distributed over a large 
number of teeth. 

The climbing, which compensates for the increase of pitch, is gradual, 
easily noticed in the running drive, does not decrease the efficiency 
of the transmission, and, as the chain lengthens and approaches- the 
top of the teeth, gives fair warning of the necessity of replacement or 
repairs of the chain. 



CHAPTER IV 

TRANSMISSION OF MOTION BY BODIES IN PURE 
ROLLING CONTACT 

83. Pure Rolling Contact consists of such a relative motion of two 
lines or surfaces that the consecutive points or elements of one come suc- 
cessively into contact with those of the other in their order. There is 
no slipping between two surfaces which have pure rolling contact, that 
is, all points in contact have the same linear speed. 

Two bodies may be rotating on their respective axes, so arranged 
that, by pure rolling contact, one may cause the other to turn with an 
angular speed bearing a definite ratio to the angular speed of the driver. 
This speed ratio may be constant or variable, depending upon the forms 
of the two bodies. The axes may be parallel, intersecting, or neither 
parallel nor intersecting * The present chapter will consider the cases of 
parallel axes connected by cylinders giving constant speed ratio, inter- 
secting axes connected by cones giving constant speed ratio and parallel 
axes connected by bodies of irregular outline, giving variable speed ratio. 

The connection between non-parallel, non-intersecting axes will be 
discussed in connection with the subject of gearing. 

84. Cylinders Rolling Together without Slipping. External Con- 
tact. In Fig. 79 let A be a cylinder fast to the shaft S and B a cylinder 
fast to the shaft Si. Assume that the shafts are held by the frame so 
that their centers are at a distance apart just equal to the sum of the radii 
of the two cylinders; that is, R + Ri = C. Then the surfaces will touch 
at P. Suppose also that the nature of the surfaces of the cylinders is 
such that, as they turn on their respective axes, there can be no slipping 
of one surface on the other. Then the surface speed of A must be equal 
to that of B, and A and B must turn in such directions relative to each 
other that the element on A which is in contact with B is moving in the 
same direction as the element on B which it touches. (Notice the 
arrows in the figure, the full arrows belonging together and the dotted 
arrows together.) 

If A makes N r.p.m. and B makes Ni r.p.m., 
Surface speed of A = 2 tRN 
and • Surface speed of B = 2 wRiNi. 

* In the case of axes which are neither parallel nor intersecting the coinciding 
elements of the rolling bodies may slide on each other in the direction of their length, 
so that the contact is not.pure rolling in a strict sense. 

63 



64 



ELEMENTS OF MECHANISM 




rtj-R.P.M 



Therefore, if the surface speed of A equals the surface speed of B, 

2 irRN = 2 irRiNi or ^ =^- (39) 

Or, in other words, the angular speeds of two cylinders which roll together 
without slipping are inversely proportional to the radii of the cylinders. 
It will be noticed that this principle is the same as that shown in the 

preceding chapter applied to 
cylinders connected by a belt 
or other flexible connector. 

85. Solution of Problems 
on Cylinders in External 
Contact. In Fig. 79 suppose 
C, N and Ni are known; re- 
quired to find the diameters 
of the two cylinders. From 
Eq. (39), 
R Nx D RiNi 

wrw or r = ~n- 

It is known also that R + Ri 
= C. R and Ri can, therefore, 
be found by solving these as 
simultaneous equations. 
The same result may also be found by a simple graphical construction 

as follows : Draw the line SS h Fig. 80, making its length equal to the 

distance between the centers of the shafts (corresponding to C, Fig. 79). 

This would, in most cases, have to 

be drawn at some reduced scale. 

From S draw a line SV, making 

any angle with SSi. From S lay off 

the distances SK equal to Ni linear 

units and KT equal to N linear units. 

The line ST is then divided into 

two parts SK and KT such that 

=t^ = -^- Now connect T with Si and from K draw a line parallel to 
Kl JS 

TS h cutting SSi at P. Then, from the similar triangles SKP and 

ST Si, J^- = §|f = §• Therefore, SP will be the radius R and PSi the 
"Si Kl JS 

radius Ri. 

Example 18. Two shafts A and B are 16 ins. on centers. A is to turn 50 times in 

a minute and B 150 times in a minute. What must be the size of the cylinders to 

connect them if they are to turn in opposite directions? 



Fig. 79 




TRANSMISSION OF MOTION 



65 



Calculation. From Eq. (39), 

Radius of A _ Turns of B per minu te 

Radius of B Turns of A per minute 

or, 

Radius of A = 3 X radius of B. 
Also 

Radius of A + radius of B = 16 ins. 
Therefore, 

Radius of B = 4 ins. 



150 
50 



and 
Radius of A = 3 X 4 



12 ins. 




ISO R..P.M. 



Graphical Solution. In Fig. 81 
draw the line A B equal to 16 ins. 
at some reduced scale. 

From A draw the line AY at 
any angle. Lay off AK equal to 

150 units. Lay off K T = 50 of the same units. Join T with B and draw KP parallel 
to TB. Then BP will be found to measure 4 ins. and AP 12 ins., making proper 
allowance for the scale at which AB was drawn. 

Example 19. A cylinder 18 ins. diameter on a shaft A making 75 r.p.m. drives by 
rolling contact a cylinder on another shaft B, the second cylinder being 4| ins. diam- 
eter. How fast does B turn if the shafts turn in opposite directions? 



r.p.m. of B _ Diam. of cylinder on A _ 18 _ 4 
r.p.m. of A Diam. of cylinder onB 4| 1 
or r.p.m. of B = 4 X r.p.m. of A = 4 X 75 = 300. 

Graphical Solution. (Fig. 82.) Draw the line AB equal in length to the sum of 

!§ + M 

2^2 



Calculation. 



the radii of the two cylinders 



llf ins. 



On AB locate the point P 9 ins. from A (therefore 2| ins. from B). From B, the 
center of the cylinder whose speed is to be found, draw a line BV at any angle and 

lay off on this line BK equal to 

B 




? R.P.M. 



75 units (that is, speed of A). 
Join K with P and through A 
draw a line parallel to Pi? cutting 
BV&tT. Then the units in KT 
will show the speed of B. 

86. Cylinders Rolling To- 
gether without Slipping. 
Internal Contact. In Fig. 
83, where the lettering corre- 
sponds to that of Fig. 79, the 
cylinder A is hollow with B inside it, so that the contact is between the 
inner surface of A and the outer surface of B. This is called internal 
contact. The same mathematical reasoning will apply here as in Fig. 
79, and Eq. (39) will hold true. The distance between centers now, 



66 



ELEMENTS OF MECHANISM 



however, is equal to R — Ri instead of R + Ri. The two cylinders in 

Fig. 83 will turn in the same direction instead of in opposite directions. 

87. Solution of Problems on Cylinders in Internal Contact. In 

Fig. 83 if C, N and Ni are known, to find the diameters of the cylinders. 
From Eq. (39), 

Ri N' 

It is also known that R — Ri = C. These may be solved as simul- 
taneous equations to find R and R\, and therefore the diameters. 

The graphical solution of prob- 
lems on cylinders rolling in 
internal contact is similar in 
principle to that shown in Fig. 80 
for external contact. Fig. 84 
shows the construction for in- 
ternal contact. 




<^ R > 

< c >p— /?,— > 

^^ 7 > 1 




Fig. 83 



Fig. 84 
y^Nj-R.P.M, 

Example 20. Two shafts A and B 
are 8 ins. on centers and are to be 
connected by rolling cylinders to turn in the same direction, A to make 20 r.p.m. 
and B to make 60 r.p.m. Find the diameters of the cylinders. 
Calculation. From Eq. (39), 

Rad. A = r.p.m. B = 60 = 3 
Rad.£ r.p.m. A 20 1 
or Rad. A = 3 X rad. B, 

also Rad. A — rad. B = 8 ins. 

3 rad. B - rad. B = 8 ins. 
or 2 rad. B = 8 ins. 

Rad. B = 4 ins. 
and Rad. A = 3 X rad. B = 3 X 4 ins. = 12 ins. 

Graphical Solution. Fig. 85 shows the graphical solution for this problem. 
Example 21. A cylinder 24 in. diameter on a shaft A making 60 r.p.m. drives 
by rolling contact a cylinder on another shaft B, the second cylinder being 6 in. 



TRANSMISSION OF MOTION 



67 



diameter. The shafts turn in the same direction, 
far apart are the shafts? 

Rev. B Diam. cyl. on A 



How fast does B turn and how 



Calculation. 



24 



Rev. A Diam. cyl. onB 6 
R.p.m. of B = 4 X r.p.m. of A = 4 X 60 



240 r.p.m. 

and Dist. between centers = rad. A — rad. B 

or Dist. between centers = 12 — 3 = 9 ins. 

Graphical Solution. Fig. 86 shows the graphical solution of Example 21. 



20 R. P. 



eoR.P.M. eQRi 




P. hi 




Fig. 85 

88. Cones Rolling Together without Slipping. External Contact. 

In the preceding discussion relating to cylinders, the shafts were neces- 
sarily parallel. It is often necessary to connect two shafts which he in 
the same plane but make some angle with each other. This is done by 
means of right cones as shown in Fig. 87 or frusta of cones as shown in 
Fig. 88, the cones having a common apex. The same reasoning applies 
to the ratio of speeds at the base of the cones as to the circles repre- 
senting the cylinders in Fig. 79. That is, 

N_ = Ri 

Ni B' 

i2i = OP sin POd and R = OP sin POC. 

Ri OP sin POCi _ sin POCi 

R 



(40) 



But 

Therefore 



OP sin POC sin POC 

Substituting this expression in Eq. (40), 



N_ 
Ni 



sin POd 



sin POC 

Therefore, the angular speeds of two cones rolling together without 
slipping are inversely as the sines of the half angles of the cones. 

89. Solution of Problems on Cones in External Contact. The law 
stated in the previous paragraph may be made use of to calculate the 
vertical angles of the cones when the angle between the axes and the 
speed ratio are known. 



68 



ELEMENTS OF MECHANISM 



Referring to Fig. 88, let angle COCi = 0, 

POC = <x and angle POCi = 0. 
N sin /3 sin /3 sin /S 



Then 



Ni sin a sin (0 — ft) sin cos 13 — cos sin 

Sin g 

Tan 5 



Cos/3 
Sin — cos 



sin /3 Sin — cos tan /3 



cos 



Whence 



Tan^ = 



Sin0 



(42) 



+ cos 



In similar manner 



Tan a 



Sin0 



+ COS0 




Fig. 87 



Graphical Construction. In Fig. 89, S and Si are two shafts which are 
to be connected by rolling cones to turn as indicated by the arrows. 
Their center lines meet at 0. S is to make iV r.p.m. and Si is to make 
Ni r.p.m. Required to find the line of contact of two cones which will 
connect the shafts, and to draw a pair of cones. 



TRANSMISSION OF MOTION 



69 



\ N-R.P.M. 




Fig. 88 




Fig. 89 



70 



ELEMENTS OF MECHANISM 



Draw a line parallel to OA, on the side toward which its direction 
arrow points, at a distance from OA equal to iVi units. Draw a similar 
line parallel to OB, N units distant from OB. These two lines intersect 
at K. A line drawn through and K will be the line of contact of the 
required cones. Select any point P on OK and from P draw lines per- 
pendicular to AO and BO meeting AO and BO at M and M h respectively. 
Produce these lines, making MH = MP and MJ = M X P. Draw HO 
and JO. Then OPH and OP J are cones of the proper relative sizes to 
connect S and Si to give the required speeds. 

If the point P had been chosen nearer to 0, the cones would have had 
smaller diameters at their bases but the ratio of the diameters would 
have been the same, or, if P had been chosen farther away from 0, the 
bases would have been larger but still of the same ratio. If frusta of 
cones are desired, the cones can be cut off anywhere, as shown by the 
dotted lines FE and FG. 

Example 22. Two shafts S and Si, Fig. 90, in the same plane, make an angle of 
105° with each other. S is to turn 90 times per minute and Si 30 times per minute. 
A cone on A having a base f ins. diameter is to roll with a cone on B to give the re- 
quired speeds; directions of rotation are to be as shown. To find the diameter of 
the base of the cone on B and to draw the cones. 



90 R. P.M. 

\ s 




Fig. 90 



Solution. Draw a line parallel to S 30 units distant from S, and a line parallel 
to Si 90 units distant from Si. These lines intersect at K; then KO is the element of 
contact. Since the base of the smaller cone is to be f ins. diameter, find a point P on 
OK which is | in. from S. Through this point draw the bases of the cones perpendic- 
ular to S and Si in the same way as explained for Fig. 89. 

Calculation of Angles a and (3. 
Using Equation (42), 

Sin0 



Tan/3 = 



+ 



where 



Tan/3 



TRANSMISSION OF MOTION 

6 = 105°, Ni = 30, N = 90, 
0.9659 



71 



12.9530. 



U + 0.2588 

|8 = 85° 25' nearly. 
2 /3 = 170° 50' nearly = angle at apex of cone on & 

a = 105° - 85° 25' = 19° 35'. 
2 a = 39° 10' = angle at apex of cone on S. 



Fig. 91 



N-R.P.M. 




N r R.P.M, 



90. Cones Rolling Together without Slipping. Internal Contact. 

The combination of speed ratio, directional relation, and angle between 
axes may be such as to require the cones to be arranged for internal 



72 



ELEMENTS OF MECHANISM 



contact. This construction is not very often met in practice, but the 
necessity for it sometimes occurs and it is well to be familiar with the 
problem. Fig. 91 shows the principle involved. Corresponding points 
are lettered the same as in Fig. 87, and the same equations apply for 
finding the ratio of speeds of the two shafts. 

It is not always possible to predict from given conditions whether the 
contact will be external or internal. 

91. Solution of Problems on Cones Rolling in Internal Contact. 
The same general methods apply to solution of problems on cones in 
internal contact as in the case of external contact. It should be noted, 
however, that since = a — /3 the equation corresponding to Equation 
(42) becomes c; in e 

Tan/3 = AT — • (43) 



N' 



cos 



Example 23. The axes of two shafts S and & intersect at an angle of 45°. S 
makes 15 r.p.m. and Si makes 60 r.p.m. The shafts are to be connected by two cones 
rolling together without slipping. The connection to be such that each turns as 
shown by the arrows. To draw the outline of the cones and calculate the angles 
at the apices. 




Fig. 92 

Graphical Solution. (Fig. 92.) Draw the center lines of the shafts, SO and SiO 
intersecting at 45° and put on the direction arrows as shown. Draw line MR 
parallel to SiO and 15 units distant from it on the side toward which the direction 



TRANSMISSION OF MOTION 



73 



arrow points. In a similar way draw TK parallel to SO and 60 units distant from it. 
The point C where MR and TK intersect will be the point through which the element 
of contact CO is drawn. Having the element of contact the. cones may be drawn as 
indicated in previous examples and are found to be in internal contact. 



Calculations of Angles of Apices. 
Using Equation (43) 



Where 



Tan /3 = 

6 = 45°, 
Tan/3 = 



Sin0 



N x = 60, 
0.7071 



AT = 15, 
0.2147. 



_. - 0.7071 

/3 = 12° 7' nearly. 2 p = 24° 14' = angle at apex of cone on Sx. 

a = 45 — 12° 7' = 32° 53' nearly. 2 a = 65° 46' = angle at apex of cone on S. 

92. Rolling Cylinder and Sphere. — Fig. 93 shows an example of a 
rolling cylinder and sphere as used in the Coradi planimeter. The seg- 
ment of the sphere A turns on an axis ac passing through a, the center 



V2 







Fig. 93 

of the sphere. The cylinder B, whose axis is located in a plane also 
passing through the center of the sphere, is supported by a frame pivoted 
at e and is held to the cylinder by a spring, not shown. The frame 
pivots e are movable about an axis at right angles to ac and passing 
through a, the center of the sphere. When the roller is in the position 
B with its axis at right angles to ac, the turning of the sphere produces 
no motion of B; when, however, the roller is swung so that its axis 
makes an angle bac\ with its former position, as shown at Z>i by dotted 
lines, the point of contact is transferred to C\ in the perpendicular from 
a to the roller axis. If now we assume the radius of the roller = R, the 
relative motion of roller and sphere, in contact at Ci, is the same as that 



74 ELEMENTS OF MECHANISM 

of two circles of radii R and bci respectively. Transferring the point of 
contact to the opposite side of ab will result in changing the directional 
relation of the motion. The action of this device is purely rolling and 
but very little force can be transmitted. It is used only in very delicate 
mechanisms. 

93. Disk and Roller. — If in Fig. 93 the radius of the sphere ac is 
assumed to become infinite and the roller B to be replaced by a sphere 
of the same diameter turning on its axis, the result will be a disk and roller 
as shown in Fig. 94, where AA represents the disk and B the roller, made 
up of the central portion of the sphere. 

If we suppose the rotation of the disk to be uniform, the velocity 
ratio between B and A will constantly decrease as the roller B is shifted 

nearer the axis of A, and 
conversely. If the roller is 
carried to the other side of 
the axis, it will rotate in the 
opposite direction to the first. 
This combination is some- 
times used in feed mechan- 
isms for machine tools, where 
it enables the feed to be 
adjusted and also reversed by 
simply adjusting the roller 
on the shaft CC. If possible 
the roller should drive. 

94. Friction Gearing. Rolling cylinders and cones are frequently 
used to transmit force, and constitute what is known as friction gearing. 
In such cases the axes are arranged so that they can be pressed together 
with considerable force, and, in order to prevent slipping, the surfaces 
of contact are made of slightly yielding materials, such as wood, leather, 
rubber or paper, which, by their yielding, transform the line of contact 
into a surface of contact and also compensate for any slight irregulari- 
ties in the rolling surfaces. Frequently only one surface is made yield- 
ing, the other being usually made of iron. As slipping is likely to take 
place in these combinations, the velocity ratio cannot be depended upon 
as absolute. 

When rolling cylinders or cones are used to change sliding to rolling 
friction, that is, to reduce friction, their surfaces should be made as hard 
and smooth as possible. This is the case in roller bearings and in the 
various forms of ball bearings where spheres are arranged to roll in 
suitably constructed races, all bearing surfaces being made of hardened 
steel and ground. 




TRANSMISSION OF MOTION 



75 




Fig. 95 



Friction gearing is utilized in several forms of speed-controlling 

devices, among which the following are good examples : 

Fig. 95 shows the mechanism of the Evans friction cones, consisting 

of two equal cones A and B turning on parallel axes with an endless 

movable leather belt C in the form of a ring running between them, 

the axis of B being urged 

toward A by means of 

springs or otherwise. By- 
adjusting the belt along 

the cones, their angular 

speed ratio may be varied 

at will. It should be 

observed that there must 

be some slipping since 

the angular speed ratio 

varies from edge to edge 

of the belt, the resulting 

ratio approaching that of the mean line of the belt. A leather-faced 

roller might be substituted for the belt and a similar series of speeds 

obtained, the cones then turning in the 
same instead of in opposite directions. 

Fig. 96 shows, in principle, another 
form, made by the Power and Speed 
Controller Co. Here two equal rollers, 
C and D, faced with a yielding material, 
are arranged to run between two equal 
hollow disks A and B. The rollers 
with their supporting yokes (only one 
of which is shown in the elevation) are 
arranged as indicated in the figure and 
are made by a geared connection, not 
shown, to turn opposite each other on 
the vertical yoke axes, s. The contour 
of the hollow in the disks must thus be 
an arc of a circle of radius equal that 
of the roller drawn from s as a center. 
If now the disk B is made fast to the 
shaft, and A, running loose, is urged 
against B by a spring or otherwise, a 

uniform motion of A may be made to give varying speeds to B by 

turning the rollers as shown. To increase the power two sets of disks 

are often used. 




Fig. 96 



76 



ELEMENTS OF MECHANISM 




Fig. 97 shows the Sellers feed disks used to give a varying angular 
speed ratio between two parallel shafts, one of them controlling the feed 
on a machine. 

The two outer wheels are thickened on their peripheries and run 
between two convex disks BB which are constantly urged together by 

hidden coil springs bearing against the 
spherical washers clearly shown. The 
disks BB are supported by the pivoted 
forked arm D. If now the disk A be 
given a uniform angular speed, the 
disk C may be made to have a greater 
or less angular speed as the axis of the 
disks BB is made to approach or recede 
from A. 

In Fig. 98 a modified form of the 
Sellers disks is shown. The shaft A 
is driven by the pulley P and is carried 
by a forked arm supplied with two 
bearings CC and swinging about a 
point near the center of the pulley 
driving P by means of a belt. The 
externally rubbing disks B are free to slide axially on the shaft A, but 
turn with it and are constantly urged apart by springs clearly shown. 
The internally rubbing convex disks are made fast to the driven shaft 
by set screws. To vary the speed of D, that of A being constant, it 
is only necessary to vary the distance between the shafts. In the 
position shown D has its highest 
speed, the disks rubbing at a. 
When the shaft A is urged in the 
direction of the arrows the rub- 
bing radius on B is diminished 
and that on E increased, the 
disks BB approaching each other. 
The disks BB may be made solid 
and one of the disks E be urged 
toward the other by a spring on 
its hub, which would simplify the 
construction. 

95. Grooved Friction Gearing. 
— Another form of friction gearing is shown in Fig. 99. Here increased 
friction is obtained between the rolling bodies by supplying their sur- 
faces of contact with a series of interlocking angular grooves; the 




TRANSMISSION OF MOTION 



77 



sharper the angle of the grooves, the greater the friction for a given 
pressure perpendicular to the axes; both wheels are usually made of 
cast iron. Here the action is no longer that of rolling bodies; but 
considerable sliding takes place, which varies with the shape and depth 
of the groove. This form of gearing is very generally used in hoisting 
machinery for mines and also for driving rotary pumps; in both 
cases a slight slipping would be an advantage, as shocks are quite 
frequent in starting suddenly and their effect is less disastrous when 
slipping can occur. 

The speed ratio is not absolute but is substantially the same as that 
of two cylinders in rolling contact on a line drawn midway between the 





Fig. 99 



Fig. 100 



tops of the projections on each wheel, they being supposed to be in 
working contact. 

96. Rolling of Non-cylindrical Surfaces. — If the angular speed ratio 
of two rolling bodies is not a constant, the outlines will not be circular. 
Whatever forms of curves the outlines take, the conditions of pure roll- 
ing contact should be fulfilled, namely, the point of contact must be on 
the line of centers, and the rolling arcs must be of equal length. For 
example, in the rolling bodies represented by Fig. 100 with o x and 
o 2 the axes of rotation, we must find the sum of the radiants in 
contact, 0\C + o^c, equal to the sum of any other pair, as o x d + o 2 e, Oif 
+ o^g; and also the lengths of the rolling arcs must be equal, cd = ce, df 
= eg. This will cause the successive points on the curves to meet on 
the line of centers, and the rolling arcs, being of equal length, will roll 
without slipping. 



78 



ELEMENTS OF MECHANISM 



There are four simple cases of curves which may be arranged to fulfill 
these conditions: 

A pair of logarithmic spirals of the same obliquity. 
A pair of equal ellipses. 
A pair of equal hyperbolas. 
A pair of equal parabolas. 

We shall also find that any of the above curves may be transformed 
in one way or another and still fulfill the conditions of perfect rolling 

contact while allowing a wide 
range of variation in the angu- 
lar speed ratio. 

97. The Rolling of two 
Logarithmic Spirals of Equal 
Obliquity. — Fig. 101 shows 
the development of a pair of 
such spirals, where, if they roll 
on the common tangent de, 
the axes Oi and o 2 will move 
along the lines fg and hk 
respectively. The arcs aic, 
cbi, etc., being equal to a 2 c, 
cb 2 , etc., and also equal to the 
distances ac, cb, etc., on the common tangent, it will be clear that if the 
axes Oi and o 2 are fixed, the spirals may turn, fulfilling the conditions of 
perfect rolling contact; for the arc cbi = arc cb 2 , and also the radiant 
0i&i + radiant o 2 b 2 = Oi'b + o 2 b = 0\C + o 2 c; and similarly for successive 
arcs and radiants. 

i 98. To construct two spirals, as in Fig. 101, with a given obliquity. — 
The equation for such a logarithmic 
spiral is 

r = ae 69 ', 

where a is the value of r when is zero; 

and 6 = : >' d> being the constant 

tan<£ 

angle between the tangent to the curve 

and the radiant to the point of 

tangency; and where e is the base of the Naperian logarithms. 

In Fig. 102 let oc = a, and ocd = <j>. Taking successive values of 9, 

starting from oc, we may calculate the values of r and thus plot the 

curve. If, however, it is desired to pass a spiral through two points on 

radiants a given angle apart, it is to be noticed from the equation of the 





TRANSMISSION OF MOTION 



79 



curve that if the successive values of are taken with a uniform increase, 
the lengths of the corresponding radiants will be in geometrical progres- 
sion. To draw a spiral through the points b and e, Fig. 102, bisect the 
angle boe, and make of a mean proportional between ob and oe; f will be 
a point on the spiral. Then by the same method bisect foe, and find 
oh; also bisect bof and find ok, and so on; a smooth curve through the 
points thus found will be the desired spiral. 

99. Continuous Motion. — Since these curves are not closed, one pair 
cannot be used for continuous motion; but a pair of such curves may be 
well adapted to sectional wheels requiring a varying angular speed. 
For example, in Fig. 103, given the axes 0\ and o 2 , the angle coie through 

which A is to turn, and the limits of the angular speed ratio. Make 



Old 



o 2 c 
equal to the minimum angular speed ratio and -^ equal to the maximum 

angular speed ratio. Then o^e must equal 0\d. Now construct a spiral 
through the points c and e. The spiral for B is that part of the spiral 





Fig. 103 



Fig. 104 



A constructed about o x which would be included between radiants 0\b 
and o x a, equal respectively to o 2 c and o 2 g (= o 2 d), which may be found 
by continuing the spiral about d beyond c or e if necessary. Since these 
curves (Fig. 103) are parts of the same spiral, and since by construction 
Oic -f o 2 c = Oie + o 2 g, A could drive B, the points e and g ultimately rolling 
together at d on the line of centers. The conditions of rolling contact are 
evidently fulfilled, as will be seen by referring to § 97. 

100. Logarithmic Spiral Driving Slide. — Fig. 104 shows a logarith- 
mic spiral sector A driving a slide B. Here the driven surface of the 
slide coincides with the tangent to the spiral, the fine of centers being 
from o through c to infinity and perpendicular to the direction of motion 
of the slide. In this combination the linear speed of the slide will equal 



80 



ELEMENTS OF MECHANISM 



the angular speed of A multiplied by the length of the radiant in contact 




101. Wheels using Logarithmic Spirals arranged to allow Complete 
Rotations. — By combining two sectors from the same or from dif- 
ferent spirals, unilobed wheels may be found which may be paired in 
such a way as to fulfill the laws of perfect rolling contact. Taking two 
j equal sectors from the same spiral, we should 
have a symmetrical unilobed wheel, as A (Fig. 
105), and this will run perfectly with a wheel B 
exactly like A, as shown. If A is the driver, the 
minimum angular speed of B will occur when the 
points d and e are in contact, and we shall have 

a.v. B _ Old 
a.v. A o 2 e 

The maximum angular speed of B will occur when 
the points / and g are in contact. Such wheels are 
readily formed, if the maximum and minimum 
angular speed ratios are known, by the method in 
§ 97, only it is to be noticed that the minimum 
ratio must be the reciprocal of the maximum 
ratio, and that the angle which each sector 
subtends must be 180° Unilobed wheels need not be formed from 
equal sectors, in which case the sectors used will not have the same 
obliquity nor will the subtended angles be equal, but the wheels must be 
so paired that sectors of the same obliquity shall be 
in contact. Fig. 106 shows a pair of such wheels 
in which maximum and minimum angular speed 
ratios occur at unequal intervals; it will, however, 
be noticed that the minimum angular speed ratio 
must here also be the reciprocal of the maximum 
ratio. 

By a similar method wheels may be formed which 
shall give more than one position of maximum and of 
minimum angular speed ratio; that is, there may be 
either symmetrical or unsymmetrical bilobed wheels, 
trilobed wheels, etc. Fig. 107 shows a pair of sym- 
metrical bilobed wheels. Here all the sectors are from 
the same spiral, all the same length, each subtending 
an angle of 90°. It will be seen that the conditions of rolling contact 
are perfectly fulfilled and that if A turns uniformly B will have two 
positions of maximum and two of minimum speed. Similarly a 




Fig. 106 



TRANSMISSION OF MOTION 



81 



pair of symmetrical trilobed wheels could be formed where each of the 
sectors subtends an angle of 60°. 

Following the method used in obtaining the unsymmetrical unilobed 
wheels of Fig. 106, a pair of unsymmetrical bilobed wheels could be 

arranged, provided only that 

sectors of the same obliquity 

come into contact and that 

such sectors subtend equal 

angles. Fig. 108 shows a pair 

of trilobed wheels of this form. 
Such wheels as those just 

described cannot be inter- 
changeable, but since any two 

spiral arcs having the same 

obliquity will roll correctly, a 

unilobe may be made to roll 

correctly with a bilobe where 

the sectors of the unilobe are 
from a given spiral and each subtending 180°, and where each of 
the sectors of the bilobe is of the same length as one of those of the 
unilobe, and from a spiral of the same obliquity, but where each 
subtends an angle of 90°. In a similar manner a trilobed wheel may be 
found which could be driven by the same unilobed wheel as above, hence 





Fig. 108 







Fig. 109 

also by the bilobed wheel found from that unilobe. These wheels would 
therefore be interchangeable. Fig. 109 shows a set of such wheels which 
would be symmetrical wheels. A set of unsymmetrical wheels could 
be found in a similar manner. 

102. The Rolling of Equal Ellipses. — If two equal ellipses, each 
turning about one of its foci, are placed in contact in such a way that the 




Fig. 110 



82 ELEMENTS OF MECHANISM 

distance between the axes Oi0 2 , Fig. 110, is equal to the major axis of 
the ellipses, we shall find that they will be in contact on the line of centers 
and that the rolling arcs are of equal length. If the point c is on the fine 
of centers Oi0 2 , we should have OiC -f- co 2 — Oi0 2 = Oxc + cd, and therefore 
cd = co 2 . Since the tangent to an ellipse at any point, as c, makes equal 
angles with the radii from the two foci, Oicm = den and ecm = o 2 cw; but 

since cd = co 2 , the point c is simi- 
larly situated in the two ellipses, and 
therefore the angle 0\cm would equal 
the angle o^cn, which would give a 
common tangent to the two curves 
at c. Hence if Oi0 2 is equal to the 
major axis, the ellipses could be in 
rolling contact on the line Oi0 2 . 
Since the distances cd and co 2 , from 
the foci d and o 2 respectively, are 
equal, it also follows that the arc 
cf is equal to the arc eg which com- 
pletes the requirements for perfect 
rolling contact. It will also be noticed that the line dee will be straight 
and that a link could connect d and e, as will be seen when discussing 
linkwork. 

If A (Fig. 110) is the driver, the angular speed ratio will vary from a 

minimum when h and k are in contact, and then equal to -^ > to a maxi- 
mum when / and q are in contact, when it will equal - 1 - • The angular 

o 2 g 

speed ratio will be unity when the major axes are parallel, the point of 
contact being then midway between 0\ and o 2 . 

Such rolling ellipses supplied with teeth, thus forming elliptic gears, 
are sometimes used to secure a quick-return motion in a slotting 
machine. 

103. Multilobed Wheels Formed from Rolling Ellipses. — Non- 
interchangeable wheels may be formed from a pair of ellipses by con- 
tracting the angles the same amount in each ellipse. Thus, if the angles 
were contracted to one-half their size, a pair of bilobed wheels could be 
formed; and if to one-third their size, a pair of trilobed wheels. Such 
wheels would give perfect rolling contact, but could only be used in pairs 
as stated. 

By a different method of contraction a pair of wheels may be formed, 
one of which may be, for example, a bilobe and the other a trilobe. By 
this method only parts of the original ellipses are used; parts which 



TRANSMISSION OF MOTION 



83 



would roll correctly, but which subtend unequal angles in some desired 
ratio. If the arcs subtend angles in proportion as 2 is to 3, the angles 
may be contracted or expanded to be 60° and 90°, which are in the same 
ratio, when we shall have arcs suitable for a trilobe and a bilobe respec- 
tively, which will roll correctly. For example, assume the foci oi and d 
(Fig. Ill) ; lay off angles foid and fde as 2 to 3. Then the point / will 
lie on an ellipse from which a bilobe and 
a trilobe may be formed by contracting 
the angle foid to 60° and the angle go 2 d 
= fde to 90°, as shown in the figure. 

104. The Rolling of Equal Parabolas. 
— Two parabolas may be considered as 
two ellipses with one focus of each re- 
moved to infinity. In the ellipses of Fig. 
110 suppose the foci 0\ and e to be so 
removed; we shall have the "parabolas of 
Fig. 112 in contact at the point c and in 
perfect rolling contact, one turning about 
its focus o 2 as an axis, and the other 
having a motion of translation perpen- 
dicular to Old. 

To prove the rolling action perfect, 
assume the parabolas with their vertexes 
in contact at m. Let / be the point on 
the turning parabola which will move to c, 
so that o 2 f = o 2 c. Draw fg parallel to o 2 c, and since the parabolas are 
equal we shall have Ig = o 2 f, therefore Ig = o 2 c; but since o 2 k is the 
directrix of the parabola whose focus is now at I, Ig = gk; therefore 
gk = o 2 c, and as this parabola slides perpendicular to o x d, the point g 
would also move to c. The rolling arcs mf and mg are equal. Thus 
the parabola turning about o 2 would cause the other parabola to have 
translation perpendicular to Old, the two moving in perfect rolling 
contact. 

105. The Rolling of Equal Hyperbolas. — If two equal hyperbolas 
are placed, as in Fig. 113, so that the distances between their foci Oi and 
o 2 , and d and e, are each equal to fg = hk, the distance between the ver- 
texes of the hyperbolas, we shall find them in contact at some point c. 
If the foci Ox and o 2 are then taken as axes of rotation, the hyperbolas will 
turn in perfect rolling contact. To prove this take the point I on the 
hyperbola whose foci are at Oi and d so that Oil = dc and 0\C = dl. Then 
since a tangent at any point on a hyperbola makes equal angles with 
the radii from the two foci, the tangent at I will bisect the angle Oild 




Fig. Ill 



84 



ELEMENTS OF MECHANISM 



and the tangent at c will bisect the equal angle c-icd. If now the branch 
oihl is placed tangent to the branch dkc with the points I and c in con- 
tact, the radius hi must fall on o^c and dl on dc. Since the difference 




Fig. 113 

between the radii from the two foci to any point on a hyperbola is a 
constant and equal to the distance between the vertexes, 0\C — dc = hk; 
but Oil was taken equal to dc, hence oic — od = hk. Then, since Oi0 2 was 



TRANSMISSION OF MOTION 85 

originally assumed equal to hk, we shall have 0\C — oj = oio 2 , and there- 
fore the line Oio 2 c will be a straight line, and the point of contact c will 
lie on the line of centers. The arc Ih which is equal to ck will also be 
equal to cf. Therefore the hyperbolas will be in perfect rolling contact. 
The same reasoning will apply for any position of the point of contact. 
It will be seen in a later chapter that since 0i0 2 = de = a constant, and 
Old = o%e = a constant, the linkage Oi0 2 ed with the axes 0\ and o 2 fixed 
would cause the same angular speed ratio about 0\ and o 2 as the rolling 
hyperbolas would give. 

If the hyperbola turning about the axis o 2 is the driver, the angular 
speed ratio will be a minimum when the vertexes / and k are in contact 

and will be -— ; this ratio will increase as the point of contact approaches 

infinity, when the ratio would be unity, and would correspond to the posi- 
tion of the linkage when 0i0 2 and de are parallel. Further rotation would 
bring the opposite branches of the hyperbolas into contact, the maxi- 
mum angular speed ratio occurring as the points g and h come together, 

when its value becomes -~ • The construction shown in the figure will 
0\h 

allow only a limited motion. 



CHAPTER V 



GEARS AND GEAR TEETH 

106. Gear Drives. It was shown in Chapter IV that one shaft 
could cause another to turn by means of two bodies in pure rolling 
contact. If the speed ratio must be exact or if much power is to be 
transmitted, a drive depending solely upon friction between the sur- 
faces of the rolling bodies is not sufficiently positive. For this reason 
toothed wheels, called gears, are used in place of the rolling bodies. 
As the gears turn the teeth of one gear slide on the teeth of the other 
but are so designed that the angular speeds of the gears are the same 
as those of the rolling bodies which they replace. 

107. Gearing Classified. In § 83 attention was called to the fact 
that rolling bodies may be used to connect axes which are parallel, 
intersecting, or neither parallel nor intersecting. The same cases 
arise in the use of gears, and special names are given to the gears 
according to the case for which they are designed. 

Gears may be classified on the above basis as follows: 

External Gears — Fig. 114. 
Internal Gears — Fig. 115 

(Here the large gear is called an annular 
and the small one a pinion) 
Twisted Spur Gear — Fig. 116. 
Herring Bone Spur Gear — Fig. 117. 
Rack and Pinion — Fig. 118. 

(The rack is a gear of infinite radius) 
Pin Gearing — Fig. 119. 
Plain Bevel (including Mitre Gears, which 

are equal bevel gears on shafts at 90°) — 

Fig. 120. 
Crown Gears — Fig. 121. 
Twisted Bevel Gears — Fig. 122. 

Hyperboloidal or Skew Gears-Fig. 123. { ^^j^*** " ^ 

dn^^r Paan'nn. f Worm and Wheel — Fig. 124. 1 Connecting Axes in 
screw rearing j Helical Gears _ Fig> 125 . J different planes. 

The name pinion is often applied to the smaller of a pair of gears. 

The various kinds of gears enumerated above will be discussed in 
more detail after the principles which apply to gearing in general 
have been considered. 



Spur Gears 



Bevel Gears 



Connecting 
Parallel 

Axes 



Connecting 
Intersect- 
ing Axes 



GEARS AND GEAR TEETH 



87 




Fig. 114 




Fig. 115 



ELEMENTS OF MECHANISM 





Fig. 116 



Fig. 117 




Fig. 118 



GEARS AND GEAR TEETH 



89 




Fig. 119 




Fig. 120 



90 



ELEMENTS OF MECHANISM 




Fig. 121 




Fig. 122 



GEARS AND GEAR TEETH 



91 




Fig. 123 




Fig. 124 



92 



ELEMENTS OF MECHANISM 



108. Speed Ratio of a Pair of Gears. It has been shown in the 
preceding chapter that if two cylinders as A and B, Fig. 126, are keyed 
to the shafts 5 and Si respectively, the angular speed of S is to the 
angular speed of Si as Z>i is to D, provided there is sufficient friction 
between the circumferences of the discs to prevent one slipping on the 
other. If the speed ratio must be exact, or if much power is to be 
transmitted, a drive like this, depending solely upon friction, is not 

positive enough. To make 
sure that there shall be 
no slipping, wheels having 
teeth around their circum- 
ferences are substituted 
for the plain discs. The 
outlines of these teeth 
must be such that the 
speed ratio is constant. 
Such a pair of wheels is 
shown in Fig. 127. Here 
the larger gear has 16 teeth 
and the smaller gear 12 
teeth. Assume that the 
shaft S is being turned 
from some external source 
of power; the gear A, 
since it is keyed to S, will 
turn with it. Then the 
teeth on A will push the 
teeth on B, a tooth on A 
coming in contact with a 
tooth on B and pushing 
that tooth along until the 
gears have turned so far around that those two teeth swing out of reach 
of each other. In order for B to make a complete revolution each one 
of its 12 teeth must be pushed along thus past the center line. There- 
fore, while B turns once 12 of the teeth on A must pass the center line 
Since A has 16 teeth in all, A will therefore make \% of a turn while B 
makes one turn. In other words, the turns of A in a given time are to 
the turns of B in the same time as the number of teeth on B are to the 
number of teeth on A , 

It is evident that the distance from the center of one tooth to the 
center of the next tooth on both gears must be alike in order that 
the teeth on one may mesh into the spaces on the other. 




Fig. 125 



GEARS AND GEAR TEETH 



93 




Fig. 126 




Tooth Face 
Addendum Tooth Flank 

Root or Ded'endum 



Fig. 127 



94 ELEMENTS OF MECHANISM 

109. Pitch Circles and Pitch Point. Let a point P (Fig. 127) be 

found on the center line SSi such that -^r = = — -=- ~ and through 

PSi Teeth on B & 

this point draw circles about S and Si as centers. Call their diameters 

D and D x . Then D = 2 PS and D x = 2 PSi. Since, as shown above, 

Revolutions of B _ Teeth on A 
Revolutions of A Teeth on B ' 
therefore, 

Revolutions of B _ D_ 
Revolutions of A D x 

That is, the two gears when turning will have the same speed ratio 
as would two rolling cylinders of diameters D and D\. The point P 
which divides the line of centers of a pair of gears into two parts pro- 
portional to the number of teeth in the gears is called the pitch point. 
The circle D, drawn through P with center at S, is the pitch circle of 
the gear A and the circle D x is the pitch circle of the gear B. 

110. Addendum and Root Circles. The circle passing through 
the outer ends of the teeth of a gear is called the addendum circle 
and the circle passing through the bottom of the spaces is called the 
root circle. 

111. Addendum Distance and Root Distance. Length of Tooth. 
The radius of the addendum circle minus the radius of the pitch 
circle is the addendum distance, or, more commonly, the addendum. 
The radius of the pitch circle minus the radius of the root circle is the 
root distance or root or dedendum. The root plus the addendum is 
the length of tooth. 

112. Face and Flank of Tooth. Acting Flank. That portion of 
the tooth curve which is outside the pitch circle is called the face of the 
tooth or tooth face. This must not be confused with the term "face 
of gear" (§ 113). The part of the tooth curve inside the pitch circle 
is called the flank of the tooth. 

That part of the flank which comes in contact with the face of the 
tooth of the other gear is called the acting flank. 

113. Face of Gear. The length of the gear tooth measured along 
an element of the pitch surface is called the length of the face of the 
gear or width of face of the gear. (See top view, Fig. 127.) 

114. Clearance. The distance measured on the line of centers, 
between the addendum circle of one gear and the root circle of the 
other, when they are in mesh, is the clearance. 

This is evidently equal to the root of one gear minus the addendum 
of the mating gear. 




GEARS AND GEAR TEETH 95 

115. Backlash. When the width of a tooth, measured on the pitch 
circle, is less than the width of the space of the gear with which it is 
in mesh, the difference between the width of space and width of tooth 
is called the backlash. This is shown in Fig. 128 where S minus T is 
the backlash, S and T being understood to be 
measured on the arcs of the pitch circles. Accu- 
rately made gears rarely have any appreciable 
amount of backlash, but cast gears or roughly 
made gears require backlash. 

116. Circular Pitch. The distance from the 
center of one tooth to the center of the next 
tooth, measured on the pitch circle, is called 
the circular pitch. This is, of course, equal to 
the distance from any point on a tooth to the 

'Ftc 1 28 

corresponding point on the next tooth meas- 

sured along the pitch circle. (See Fig. 127.) The circular pitch is 
equal to the width of tooth plus the width of the space between teeth, meas- 
ured on the pitch circle. The whole circumference of the pitch circle 
is equal to the circular pitch multiplied by the number of teeth, or 
the circular pitch is equal to the circumference of the pitch circle 
divided by the number of teeth. In Fig. 127 let T represent the 
number of teeth in the gear A and let C represent the circular pitch. 
Then, 

C= T f- (44) 

Two gears which mesh together must have the same circular 
pitch. 

117. Diametral Pitch and Pitch Number. Module. The term 
diametral pitch is used by different authorities to mean two different 
quantities. Most gear makers' catalogues and many books on gearing 
define diametral pitch as the number of teeth per inch of diameter of 
pitch circle. For example, if a gear has 24 teeth and the diameter of 
its pitch circle is 8 in., these authorities would say that the diametral 
pitch of the gear is 24 divided by 8 or is 3. Such a gear is described 
as a 3-pitch gear. This is also called the pitch number. 

Other authorities define diametral pitch as the length of pitch 
diameter which the gear has per tooth, or the ratio of the diameter 
to the number of teeth. That is, referring again to the 24 tooth gear 
having a pitch diameter of 8 in., according to this latter definition the 
diametral pitch would be 8 in. divided by 24 or § of an inch. Another 
name for this quantity is module. 



96 ELEMENTS OF MECHANISM 

Throughout this book the terms diametral pitch and pitch number 
will be used interchangeably, meaning the number of teeth per inch 
of pitch diameter, and the name module will be used for the amount 
of diameter per tooth. 

If M represents the module and P.N. the pitch number or diametral 
pitch, T the number of teeth and D the pitch diameter, the above 
may be expressed in the form of equations as follows : 

(45) 

(46) 





Pitch diameter D 
M ~ Teeth ~ T 




p v Teeth _. T 




Pitch diameter D 


Therefore, 


M =A.- 


118. Relation between Circular Pitch and Mo 


From Eq. (45) 


*-?. 


and from Eq. (44) 


ttD 

c- T • 


Dividing Eq. (44) by Eq. (45), 




C irD . D 

M T ' T T ' 


or 


C = MXir, 


or 


L P.N. 



(47) 

(48) 

Or, in words, the circular pitch is equal to the module multiplied by t. 

119. Angle and Arc of Action. The angle through which the driv- 
ing gear turns while a given tooth on the driving gear is pushing the 
corresponding tooth on the driven gear is called the angle of action 
of the driver. Similarly, the angle through which the driven gear 
turns while a given one of its teeth is being pushed along is called the 
angle of action of the driven gear. The angle of approach, in each 
case, is the angle through which the gear turns from the time a pair 
of teeth come into contact until they are in contact at the pitch point. 
It will be shown later that the pitch point is one of the points of con- 
tact of a pair of teeth during the action. The angle of recess is the 
angle turned through from the time of pitch point contact until con- 
tact ceases. 

The angle of action is therefore equal to the angle of approach, 
plus the angle of recess. 



GEARS AND GEAR TEETH 97 

In Fig. 129 a tooth M on the driving gear is shown (in full lines) just 
beginning to push a tooth N on the driven gear. The dotted lines 
show the position of the same pair of teeth when N is just swinging out 
of reach of M. While M has been pushing N, any radial line on the 
gear B, as, for example, the line drawn through the center of the tooth 
M , has swung through the angle K, and any line on gear A has swung 
through the angle V. K is, there- 
fore, the angle of action of the gear 

B, and V is the angle of action of _____ -^W*-? K \ Dr!ver 

the gear A. 

It should be noted that the 
angles of approach and recess are 
not shown in Fig. 129. 

The arc of action is the arc of the 
pitch circle which subtends its angle of action. The arcs of approach 
and recess bear the same relation to the angles of approach and recess 
as the arc of action bears to the angle of action. Since the arcs of 
action on both gears must be equal, the angles of action must be 
inversely as the radii. 

Therefore the following equation holds true: 

Angle of action of driver _____ Number of teeth on driven gear .. . 
Angle of action of driven gear Number of teeth on ^ 




The arc of action must never be less than the circular pitch, for, if it 
were, one pair of teeth would cease contact before the next pair came 
into contact. 

120. The Path of Contact. Referring still to Fig. 129, the teeth 
as shown in full lines are touching each other at one point a. This 
point is really the projection on the plane of the paper of a line of con- 
tact equal in length to the width of the gear face (see § 113). In the 
position shown dotted the teeth touch each other at the point b. If 
the teeth were drawn in some intermediate position, they would touch 
at some other point. For every different position which the teeth 
occupy during the action of one pair of teeth they have a different 
point of contact. A line drawn through all the points at which the 
teeth touch each other (in this case the line aPb) is called the path of 
contact. This may be a straight line or a curved line, depending upon 
the nature of the curves which form the tooth outlines. In all properly 
constructed gears the pitch point P is one point on the path of contact. 

121. Obliquity of Action or Pressure Angle. The angle between the 
line drawn through the pitch point perpendicular to the line of centers, 
and the line drawn from the pitch point to the point where a pair of 



98 ELEMENTS OF MECHANISM 

teeth are in contact is called the angle of obliquity of action or pressure 
angle. In some forms of gear teeth this angle remains constant while 
in other forms of teeth it varies. 

The direction of the force which the driving tooth exerts on the 
driven tooth is along the line drawn from the pitch point to the point 
where a pair of teeth are in contact (see § 122). The smaller the 
angle of obliquity the greater will be the component of the force in 
the direction to cause the driven gear to turn and the less will be the 
tendency to force the shafts apart. In other words, a large angle of 
obliquity tends to produce a large pressure on the bearings. 

122. Law Governing the Shape of the Teeth. The curves which 
form the outline of the teeth on a pair of gears may, in theory at least, 
have any form whatever, provided they conform to one law, namely: 
The line drawn from the pitch point to the point where the teeth are in 
contact must be perpendicular to a line drawn through the point of con- 
tact tangent to the curves of the teeth. 

That is, the common normal to the tooth curves at all points of contact 
must pass through the pitch point. 

This is illustrated in Fig. 130. The teeth in the full line position 
touch each other at a. That is, the curves are tangent to each other 




at this point. The line ST is drawn tangent to the two curves at a. 
The curves must be made so that this tangent line is perpendicular 
to the line drawn from a to P. Similarly, in the dotted position the 
line VW which is tangent to the curves at their point of contact b 
must be perpendicular to the line bP. This must hold true for all 
positions in which a pair of teeth are in contact, in order that the 
speed ratio of the gears shall be constant. 

This law may be proved as follows: Let An and Bm be lines drawn 
from the centers A and B perpendicular to the common normal through 



GEARS AND GEAR TEETH 99 

the point of contact a. Let co A = angular speed of gear A (expressed 
in radians) and <a B = the angular speed of B. Then linear speed of 
n = 03 A X An and linear speed of m = o) B X Bm. The direction of 
motion of m and n are both along nm at the instant under consideration 
and the motion is the same as if wheel A were pulling B by an inexten- 
sible cord attached at m and n. That is, the linear speed of m = linear 

speed of n. Therefore oi A X An = oi B X Bm or— = -= — But since 

the angular speeds of the radii to two points which have the same linear 

speed are inversely as the radii, — = -j~5' 

COb A.JT 

Therefore, 25? "IF 

Hence raw must intersect the line of centers AB at P. 

123. Conjugate Curves. Two curves are said to be conjugate 
when they are so formed that they may be used for the outline of 
two gear teeth which will work on each other and fulfill the law 
described in § 122. 

124. To Draw a Tooth Outline which shall be Conjugate to a Given 
Tooth Outline. Given the face or flank of a tooth of one of a pair of 
wheels, to find the flank or face of a tooth of the other. The solution 
of this problem depends on the fundamental law, § 122. In Fig. 131 
let the flank and face of a tooth on A be given. If A is considered as 
the driver, points on the flank, as a and 6, will be points of contact in 
the approaching action, and by the law they can properly be points 
of contact only when the normals to the flank at these points pass 
through the pitch point; therefore drawing ac and bd normals to the 
flank from the points a and b respectively, and then turning A back- 
ward until the points c and d are at the pitch point, we find positions 
a,i and 6i which a and b respectively must occupy when they can be 
points of contact with the face of a tooth of the other wheel. The 
point ai must be a point on the desired face of a tooth on the wheel B 
when the pitch circles have been moved backward an arc equal to 
CiC, that is, so that c is at the pitch point. To find this point when 
the teeth are in the original position, it is necessary to move the wheels 
forward, the wheel B carrying with it the point «i and the normal 
aiCi until the point C\ has moved through an arc C\C% equal to C\c; this 
will carry the point a\ to (h, and the normal a\C\ to a^d. During this 
same forward motion the normal a^d moving with the wheel A will 
return to its original position ac. 

In a similar manner the point 61, which can be a point of contact of 



100 



ELEMENTS OF MECHANISM 

Oik 




o4> 

Fig. 131 



GEARS AND GEAR TEETH 



101 



the given flank with the desired face, is a point on this face when the 
pitch circles of the wheels are moved backward an arc equal to Cid. 
Moving them forward the same distance, the point 61 and normal biCi, 
moving with the wheel B, will be found at b 2 and b 2 d 2 . This process 
may be continued for as many points as may be needed to give a 
smooth curve. The curve drawn through the points a^ci will be 
the required face. 

A similar process gives the flank of the tooth on the wheel B which 
will work properly with the given face. The normals taken in the 




Fig. 132 

figure are eg and fh, the positions of e and / when they can be points 
of contact being e\ and f x ; and the points on the required flank when 
in the original position are e% and / 2 . 

A smooth curve passed through the points of contact ai&iCiei/i will 
be the path of contact, the beginning and end of which will be deter- 
mined by the addendum circles of B and of A respectively. 

Fig. 132 shows the same construction for finding the curve for a 
pinion tooth conjugate to a given rack tooth. 



102 



ELEMENTS OF MECHANISM 



Another method of solving the above problem is shown in Fig. 133, 
where Oi and o 2 are a pair of plates whose edges are shaped to arcs of 
the given pitch circles AA X and BB h due allowance being made for a 
thin strip of metal, gh, connecting the plates, to insure no slipping of 
their edges on each other. 

Attach to o 2 a thin piece of sheet metal, M, the edge of which is 
shaped to the given curve aa; and to oi a piece of paper, D, the piece 
M being elevated above Oi to allow space for the free movement of D. 

Now roll the plates together, keep- 
ing the metallic strip gh in tension, 
and, with a fine marking-point, 
trace upon the paper D, for a suffi- 
cient number of positions, the out- 
line of the curve aa x . A curve just 
touching all the successive outlines 
on D, as eei, is the corresponding 
tooth curve for 0\. 

125. To Draw the Teeth of a 
Pair of Gears. When the tooth 
outlines have been found, and 
the circular pitch, backlash, ad- 
dendum, and clearance are known, 
the teeth may be drawn as 
shown in Fig. 134. Let MN and 
RT be the known tooth outlines 
for the gears A and B respectively, 
gear, one pair of which shall be 
in contact at the pitch point. Assume no backlash and the width 
of the teeth equal to the spaces on the pitch circles. Draw the ad- 
dendum circles of each with radius equal to radius of pitch circle 
plus addendum. Draw root circle of each with radius equal to radius 
of pitch circle minus (addendum of other gear plus clearance) . Space 
off the circular pitch on either side of P on each pitch circle. This 
may be conveniently done by drawing a fine tangent to the pitch circles 
at P, laying off the circular pitch PC and PC\ on this line. Set the 
dividers at some small distance such that when spaced on the pitch 
circles the length of arc and chord will be nearly the same. Start at C, 
step back on CP until the point of the dividers comes nearly to P (say 
at K) then step back on the pitch circles the same number of spaces, 
getting H and L. Hi and Li can be found in the same manner. 

Through the point J, where the curve RT cuts the pitch circle of B, 
draw the radial line cutting the addendum circle at V. Make arc WX 




To draw three teeth on each 



GEARS AND GEAR TEETH 



103 



equal to arc VR. Cut a templet or find a place on a French curve which 
fits the curve RJT, mark it, and transfer the curve to pass through X 
and P. Make PH 2 equal to one half PH h turn the curve over and draw 




Fig. 134 

curve through H 2 in the same way that PX was drawn. All the other 
curves may be drawn in a similar way. 

126. Clearing Curve. If the flanks are extended until they join 
the root line, a very weak tooth will often result; to avoid this, a 



104 



ELEMENTS OF MECHANISM 



fillet is used which is limited by the arc of a circle connecting the root 
line with the flank, and lying outside the actual path of the end of 
the face of the other wheel. This actual path of the end of the face 
is called the true clearing curve. 




This curve is the epitrochoid traced by the outermost corner of one 
tooth on the plane of the other gear. The general method of drawing 
such a curve is shown in Fig. 135. The tooth M is to work in the space 
N. From e lay off the equal arcs ee h eie 2 , e 2 e 3 , etc., and from / lay off 



GEARS AND GEAR TEETH 



105 



the same distance //i, jji, etc. From /, /i, / 2 , etc., draw arcs with the 
radii eR, eji, e 2 R, etc., respectively. A smooth curve internally tangent 
to all these curves will be the desired epitrochoid or clearing curve. 

127. The Involute of a Circle. The form of the curve most com- 
monly given to gear teeth is that known as the involute of a circle. 
Teeth properly constructed with this curve will conform to the law 
described in § 122 as will appear in the following paragraphs. This 
curve and the method of drawing it will, therefore, be studied before 
considering the method of applying it to gear teeth. 

In Fig. 136 the circle represents the end view of a cylinder around 
which is wrapped an inextensible fine thread, fastened to the cylinder 
at A and having a pencil in a loop at P. If now the pencil is swung 
out so as to unwind the thread 
from the cylinder, keeping it 
always taut, the curve which the 
pencil traces on a piece of paper 
on which the cylinder rests is 
known as an involute of the 
circle which represents the end 
view of the cylinder. The same 
result is obtained by considering 
the tracing point to be carried 
by a line rolling on a circle. All 
involutes drawn from the same 
circle are alike, but involutes 
drawn from circles of different 
diameters are different. The 
greater the diameter of the circle 
the flatter will be its involute. 

In constructing the involute 
of a circle on the drawing board 
it is, of course, impossible actu- 
ally to wrap a thread around 
the circle and draw the involute 
by unwinding the thread. Fig. 
137 shows the method of constructing an involute on the drawing 
board. Suppose the involute is to be drawn starting from any point p 
on the circle whose center is (7. Set the dividers at any convenient short 
spacing; a distance which is about f± the circumference of the circle 
will give good results. Place one of the points of the dividers at p and 
space along on the circumference a few times, getting the equidistant 
points m, n, r, s. At each of these points draw radial lines and con- 




106 



ELEMENTS OF MECHANISM 



struct lines perpendicular to these radii as shown. Each of these 
perpendiculars will then be tangent to the circle at one of the points. 
Taking care that the setting of the dividers remains unchanged, lay 
off one space ml on the tangent at m. On the next line, which is 
tangent at n, lay off from n the same distance twice, getting the point 
2. From r lay off the distance three times, getting the point 3; and so 
on until points are found as far out as desired. A smooth curve drawn 
through these points with a French curve will be a very close approxi- 
mation to the true involute, — close enough for all practical purposes 
if the work is done carefully. 

128. Application of the Involute to Gears. In Fig. 138 let A and B 
be the centers of two gears whose pitch circles are tangent at P. Through 




Pitch Circle 
\\ 
""■"-- -Base Circle 



V 



Fig. 138 



P draw a line XX perpendicular to the line of centers AB and another 
line YY making an angle K with XX. From A draw a line Aa perpen- 
dicular to YY and from B draw Bb also perpendicular to YY. Then A a 
and Bb will be the radii of circles drawn from A and B respectively, tan- 
gent to YY. These circles are called base circles. The triangle AaP is 

similar to the triangle BbP, therefore, -m = ttfj • That is, the radii of 
& ' .' AP BP ' 

the base circles are in the same ratio as the radii of the pitch circles. 

Therefore, since 

Angular speed of A _ RP 

Angular speed of B AP 



GEARS AND GEAR TEETH 107 

it follows that Angular speed of A = Bb 

Angular speed of B Aa 

If now the tooth outlines on the gear A are made involutes of the 
circle whose radius is A a and those on B involutes of the circle whose 
radius is Bb, a tooth on A will drive a tooth on B in such a way that at 
all times the angular speed of A will be to the angular speed of B as 
Bb is to Aa, the action being the same as if the lines ab and aA were 
inextensible cords connecting the base circles and the involutes were 
curves traced by marking points on the cords. The same ratio of speeds 
would hold if B were the driver. The teeth would always be in contact at 
a point on the line aPb or at a point on aiPbi. The path of contact in 
gears having involute teeth is, therefore, a straight line and the angle of 
obliquity or pressure angle is constant. That is, the direction of the force 
which the driving tooth exerts on the driven tooth is the same at all times. 

129. To Draw a Pair of Involute Gears. Suppose that it is required 
to draw a pair of involute gears 4-pitch, 16 teeth in the driver and 12 
teeth in the driven gear; addendum on each to be \ in. and dedendum 
■£% in. ; pressure angle 6 = 22\°. 

One-half of one gear and 4 teeth on the other will be drawn. In 
Fig. 139 draw a center line and on this line choose a point S which is 
to be the center of the driving gear. To find the distance between 
centers and thus locate the center of the other gear, first find the pitch 
diameter of each. Since the driver has 16 teeth and is 4-pitch (that is, 
it has 4 teeth for every inch of pitch diameter), its pitch diameter must 
be 16 -T- 4 or 4 in. In like manner the diameter of the other gear is 
12 -r- 4 or 3 in. The distance between centers must be equal to the 
radius of the driver plus the radius of the driven gear and is, therefore, 
2 + 1| in. or 3J in. Measure off the distance SSi equal to 3| in. and 
Si is the center of the driven gear. Next, locate the pitch point P, 2 in. 
from S or 1| in. from Si, and through P draw arcs of circles with S and 
Si as centers. These arcs are parts of the pitch circles of the two gears. 
Through P draw the line XX perpendicular to the line of centers and 
draw the fine YY making an angle of 22|° with XX. From S and Si 
draw lines perpendicular to YY meeting it at a and b. With radii Sa 
and SJ) draw the base circles. Draw the addendum circle of the upper 
gear with S as a center and radius equal to the radius of the pitch circle 
plus the addendum distance. This will be 2\ in. In similar manner 
draw the addendum circle of the lower gear with a radius If in. Draw 
the root circle of the upper gear with Sasa center and radius 2—^2 
in. (that is pitch radius — dedendum). Find the root circle of the lower 
gear in a similar way. We are now ready to construct the teeth. 



108 



ELEMENTS OF MECHANISM 



From the point a space off on the base circle the arc at equal in length to 
the line aP and from t, thus found, draw the involute of the base circle 
of the upper gear as described for Fig. 137. tPk is the curve thus found. 
In a similar manner find the point r such that arc br is equal in length 
to the line bP and from r draw the involute rPn of the lower base circle. 

The shape of the tooth curves having been found in this way, the 
next step is to find the width of the teeth on the pitch circles and draw 
in the remaining curves. Since the gears are 4-pitch, the circular pitch 




is | X 3.1416 = 0.7854 in. and if the width of the tooth is one-half the 
circular pitch, as is usually the case, the width of the tooth on each gear 
must be | X 0.7854 or 0.39 in. nearly. Therefore, lay off the arc PR 
equal to 0.39 in. and through R draw an involute which is a duplicate 
of the curve tPk except that it is turned in the reverse direction. Simi- 
larly make PT equal 0.39 and draw an involute through T which is a 
duplicate of the curve nPr. These curves can be transferred to the new 
positions by means of templets, it being unnecessary to construct the 
curve more than once. The part of the tooth outlines below the base 



GEARS AND GEAR TEETH 



109 



circles may be made radial lines with small fillets at the bottom corners. 
One tooth on each gear has now been completed and other teeth may- 
be drawn like these by means of templets. 

If the larger gear is the driver and turns in the direction indicated by 
the arrow, the path of contact is the fine MPN. 




Fig. 140 



130. Normal Pitch. The normal pitch is the distance from one 
tooth to the corresponding side of the next tooth, measured on the com- 
mon normal (CCi, Fig. 140) . From the method of generating the curves 
this distance is constant and is equal to the distance between the corre- 
sponding sides of two adjacent teeth measured on the base circle (arc 
KK h Fig. 140). 



110 ELEMENTS OF MECHANISM 

The definite quantities in a given involute gear are the base circle 
and the normal pitch. 

131. Relation between Normal Pitch and Circular Pitch. Referring 
to Fig. 140, let D represent the diameter of the pitch circle and D b the 
diameter of the base circle. N = normal pitch, C = circular pitch, 
T the number of teeth, a the point where the line of obliquity (or generat- 
ing line) is tangent to the base circle, and the pressure angle. Draw 
Sa and produce it to meet XX (the tangent to the pitch circle through 
P) at W. Angle aSP = angle aPW = 0, and the triangles aPW and 
aSP, are similar. 

Therefore, -^ = cos 0. 

From the definition of normal pitch (§ 130) 

N = 
and from equation (44) 



ttD 6 
T 

ttD 



c T 

Therefore, ^ = j? = cos 0. (50) 

That is, the normal pitch is equal to the circular pitch multiplied by the 
cosine of the pressure angle. 

132. Relation between Length of Path of Contact and Length of 
Arc of Contact. In Fig. 141 the teeth shown in full lines are in con- 
tact at the beginning of the path of contact and the teeth shown 
dotted are at the end of the path of contact. 

The angle a is therefore the angle of action on the driven gear and 
arc NPM is the arc of action. The corresponding arc of the base circle 
is LK. 

As previously shown, 

Radius base circle _ 
Radius pitch circle 

Therefore, ^7^ = cos 6. 

' arc NM 

But, from the properties of the involute, considering the line ab as the 
connecting line between the two revolving base circles, the length of 
the line TT\ (that is, the path of contact) is equal to the length of the 
arc LK. 

T T 
Therefore, arc ^r^ = cos 6. (51) 



GEARS AND GEAR TEETH 



111 



That is, the path of contact is equal to the arc of contact multiplied by the 
cosine of the pressure angle. 

A convenient method for finding this relation graphically is as follows : 
Draw XX tangent to the pitch circles at P. From T and T x draw TR 




Driven 



Fig. 141 



and TiS perpendicular to TT h meeting XX at R and S. The length 
RS is then equal to the length of the arc of action, RP being the length 
of the arc of approach and PS the length of the arc of recess. 



112 



ELEMENTS OF MECHANISM 



Conversely, if the lengths of the arcs of approach and recess are 
known, the ends of the paths of contact may be found by laying off the 
arcs along XX and drawing perpendiculars to ab. 

133. Limits of Addendum on Involute Gears. Fig. 141 shows one 
tooth on each of a pair of 4-pitch gears of 18 and 24 teeth respectively. 




The addendum arcs of the teeth, shown in full lines, are such that the 
addendum distance is equal to f the module (^ m 0- If f° r an y reason 
it is desired to redesign these gears with longer teeth, that is, with larger 



GEARS AND GEAR TEETH 



113 



addendum circles, it will be necessary to know how long the teeth can 
be made without causing trouble. The tooth on B can be increased in 
length until the addendum circle passes through the point a, where the 
line of obliquity YY is tangent to the base circle of A. If the tooth is 
made longer than this limit, interference will result unless some special 
form of curve is constructed in place of the involute for the outer end 
of the tooth. 

At the right of Fig. 142 are shown the same pair of teeth with the 
addendum of gear B lengthened so that the addendum circle is outside 
of point a. It will be noticed that the extended face of the tooth 




Fig. 142a 

of B cuts into the radial extension of the flank of the tooth on A 
and also cuts slightly into that part of the tooth outside the base 
circle. 

To show that interference of this sort occurs refer to Fig. 142a where 
a 22-tooth gear B is shown working with a 10-tooth pinion A. P is the 
pitch point, a the point where the base circle of the pinion is tangent 
to the line of obliquity. If the addendum of the gear B is carried beyond 
a (say, to k) the conjugate to the line ak is the involute na of the pinion 
base circle and lies in the space between the pinion teeth. To see that 
this is true assume that the line of obliquity is a cord carrying a point 
which traces the involutes (see § 128) and, instead of wrapping around 
the base circle of the pinion, is tangent to it at a and extends beyond 



114 ELEMENTS OF MECHANISM 

carrying a marking point at n x . As the gear B turns to the left the cord, 
by its contact at a, turns the base circle of the pinion right-handed, 
the tracing point n\ traces the outline k\a y (same as ka) on the plane of 
the gear B and the curve n x a x (same as no) on the plane of the pinion A. 

Next, to show that the curve kxa v will cut into curve aA (the regular 
pinion tooth) as the tracing point approaches a, assume the tracing 
point to be at n 2 . The curve k 2 a w (same as ka) will be tangent to a 2 n w 
(same as an) at n 2 . Now adz is tangent to a 2 n w at a%. Therefore, since 
kza w and adz are tangent to the same curve a 2 n w at different points, their 
directions must be such that they tend to intersect. In the figure the 
interference is evident. When the tracing point reaches a the inter- 
ference ceases. 

The above consideration shows that while the common normal will 
trace conjugate curves beyond a, it is impossible to get proper action 
for that portion of the tooth outline kar outside of a, first because the 
conjugate to ka lies inside the tooth of which kar is the outline, and sec- 
ond because the part of the tooth of which ka is the outline cuts into 
the tooth outline at. 

Referring again to Fig. 142, the tooth on A might be lengthened until 
the addendum circle passed through the point of tangency b except for 
the fact that there is another limit to the addendum which sometimes 
has to be considered. The maximum addendum here is limited by the 
intersection of the two sides of the tooth giving a pointed tooth. It is 
evident that no further increase in addendum is here possible. 

The following illustration will help to make the above statements 
clear. 

In Fig. 143 let it be required to determine if the arc of recess can 
be equal to f of the circular pitch. Lay off from a on the tangent the 
distance ab = f of the circular pitch. Draw be perpendicular to the 
line of obliquity; c will be the end of the path of contact for the given 
arc of recess. If the point c came beyond d, the tangent point of the 
line of obliquity and the base circle, the action would be impossible since 
no contact can occur beyond d. But if, as in Fig. 143, the point c comes 
between a and d, it is necessary to determine if the face of the tooth 
on A can reach to c. Lay off on the pitch circle A the arc ae = ab = f 
of the pitch; the face of the tooth on A will then pass through c and e. 
Draw the line co from c to the center of A , and note the point / where it 
cuts the pitch circle A. If ef is less than one-half the thickness of the 
tooth, the action can go as far as c and the teeth will not be pointed. 
In the figure, assuming tooth and space equal, the thickness of the tooth 
would be eg, and ef is less than \ eg) therefore the action is possible, as 
is shown by the two teeth drawn in contact at c. 



GEARS AND GEAR TEETH 



115 




Fig. 143 



116 



ELEMENTS OF MECHANISM 



134. Involute Pinion and Rack. Fig. 144 shows a pinion driving 
a rack. The path of contact cannot begin before the point a, but the 
recess is not limited excepting by the addendum of the pinion, since 
the base line of the rack is tangent to the line of obliquity at infinity. 
For the same reason it will be evident that the sides of the teeth of the 
rack will be straight lines perpendicular to the line of obliquity. In the 
figure the addendum on the rack is made as much as the pinion will 
allow, that is, so that the path of contact will begin at a. The adden- 
dum of the pinion will give the end of the path of contact at 6. 




Fig. 144 

In Fig. 145, the diagram for a pinion and a rack, let it be required 
to determine if the path of contact can begin at a and go as far as b; 
to be solved without using the tooth curves. For the contact to begin 
at a the face of the rack must reach to a. Draw the line ac perpendicu- 
lar to the line of obliquity, giving cd as the arc of approach; draw ae 
parallel to the line of centers, and if ce is less than one-half the thick- 
ness of the rack tooth, the approaching action is possible without 
pointed teeth. Similarly for the recess, draw the line bf perpendicular 
to the line of obliquity, giving df equal to the arc of recess; make the 
arc dg on the pinion's pitch circle equal to df, then the face of the pinion's 
tooth will pass through b and g; draw the line bh to the center of the 
pinion, and note the point h where it crosses the pinion's pitch circle. 
If gh is less than one-half the thickness of the tooth, the recess is pos- 
sible without pointed teeth. 



GEARS AND GEAR TEETH 



117 



135. Involute Pinion and Annular Wheel. Fig. 146 shows an invo- 
lute pinion driving an annular wheel. This case is very similar to a 
pinion and rack. The addendum of the annular is limited by the tan- 
gent point a of the pinion's base circle and the line of obliquity, while 
the addendum of the pinion is unlimited except by the teeth becoming 
pointed. The base circle of the annular lies inside the annular, so that 
its point of tangency with the line of obliquity is at 6. If we take some 
point on the line of obliquity, as c, and roll the tooth curves as they 
would appear in contact at that point, the teeth of the annular will be 




Fig. 145 



found to be concave, and the addendum of the annular would seem to be 
limited by the base circle of the annular where the curves end. But if 
these two teeth are moved back until they are in contact at a, it will be 
evident that the annular's tooth curve cannot be extended beyond a 
without interfering with the pinion teeth as in the case of the gear in 
Fig. 142a. Therefore the addendum of the annular is limited by the 
point of tangency of the base circle of the pinion and the line of obliquity. 
If the ratio of the number of teeth in the pinion to the number of 
teeth in the annular exceeds a certain limit, interference will occur 



118 



ELEMENTS OF MECHANISM 



between the teeth after they have ceased contact along the path of con- 
tact. This is illustrated in Fig. 146a, where an 18-tooth pinion is shown 




Fig. 146 



with a 24-tooth annular of 14|° obliquity. Their teeth are shown inter- 
fering at K. 

The limiting size of the pinion at which this interference begins to be 
evident is a function of the angle of obliquity. 



GEARS AND GEAR TEETH 



119 



136. Possibility of Separating two Involute Wheels. Interchange- 
able Gears. One of the most important features of involute gearing 
is the fact that two such wheels may be separated, within limits, without 
destroying the accuracy of the angular speed ratio. In this way the 
backlash may be adjusted, since the original pitch circles need not be 
in contact. To show that this is so, the gears shown in Fig. 147 may 
be redrawn using the same pitch circles and base circles, but separating 
them slightly, keeping the teeth in contact, as has been done in Fig. 
148. Connect the base circles by the tangent be. If now the line be 
carries a marking-point, it will evidently trace the involutes of the two 
base circles, as de and he, and these curves must be the same as the tooth 




Fig. 146a 



curves in Fig. 147. In Fig. 148 these curves de and he will give an 
angular speed ratio to the base circles inversely as their radii, but the 
radii of these base circles are directly as the radii of the original pitch 
circles (Fig. 147) ; hence in Fig. 148 the tooth curves de and he would 
give an angular speed ratio to the two wheels inversely as the radii of 
the original pitch circles, although these circles do not touch. The path 
of contact is now from k to e, which is considerably shorter than in 
Fig. 147; it is, however, slightly more than the normal pitch, so that the 
action is still sufficient. The limit of the separation will be when the 
path of contact is just equal to the normal pitch. The pressure angle 
is bam, which is greater than in Fig. 147. The backlash has also 
increased. 



120 



ELEMENTS OF MECHANISM 



The wheels have new pitch circles in contact at a, and a new angle 
of obliquity or pressure angle, also a greater circular pitch with a certain 




Fig. 147 



amount of backlash; and if these latter data had been chosen at first 
the result would have been exactly the same wheels as in Fig. 147, slightly 
separated. It will be seen that the radii of the new pitch circles are to 



GEARS AND GEAR TEETH 



121 



each other as the radii of the respective base circles, and consequently 
as the respective original pitch circles. It will also be seen that the 




line of obliquity, which is the common normal to the tooth curves, passes 
through the new pitch point a so that the fundamental law of gearing 
is still fulfilled. 



122 ELEMENTS OF MECHANISM 

By the application of the preceding principles two or more wheels 
of different numbers of teeth, turning about one axis, can be made to 
gear correctly with one wheel or one rack; or two or more parallel racks 
with different obliquities of action may be made to gear correctly with 
one wheel, the normal pitches in each case being the same. Thus dif- 
ferential movements may be obtained which are not possible with teeth 
of any other form. 

In this same connection, attention may be called to the fact that in 
a set of involute gears which are to be interchangeable the normal pitch 
must be the same in all. 

137. Standard Proportions. There is no one standard governing 
the relations between pitch, addendum, clearance, etc. Two methods 
of proportioning the teeth may be mentioned which, for convenience, 
will be referred to as the Brown & Sharpe Standard and the American 
Society of Mechanical Engineers Standard. The Brown & Sharpe 
standard represents the proportions ordinarily used by the Brown & 
Sharpe Manufacturing Company. Their practice is to modify the form 
of the tooth curves slightly at the end to avoid interference, or as expe- 
rience has shown them to be desirable. The A.S.M.E. standard is that 
proposed in a majority report of a committee appointed to recommend 
a standard which would be desirable for general adoption. 

The following tables give the proportion for the two standards. 

Table I. — BROWNE & SHARPE STANDARD FOR 
INVOLUTE GEARS 

Angle of obliquity (pressure angle) 14£° 

Addendum Equal to module 

Clearance Approximately | module 

Dedendum or root Approximately If module 

Table II. — A.S.M.E. STANDARD (Proposed) FOR 
INVOLUTE GEARS 

Angle of obliquity (pressure angle) 22|° 

Addendum J module 

Clearance § module 

Dedendum or root Equal to module 

Another standard, differing slightly from the A.S.M.E. standard, 
gives the form of tooth known as the stub tooth. 

138. Cycloidal Gears. Formerly, gear teeth were constructed on 
the cycloidal system. The faces of the teeth were epicycloids generated 
on the pitch circles and the flanks hypocycloids generated inside the 
pitch circles. The involute system has replaced the cycloidal almost 



GEARS AND GEAR TEETH 



123 



entirely for general purposes, although cycloidal teeth are still used in 
some special cases. 

In Fig. 149 let 0\ and o 2 be the centers of the two wheels A and B, 
their pitch circles being in contact at the point a. Let the smaller circles 
C and D, with centers at pi and p 2 , be placed so that they are tangent 
to the pitch circles at a. Assume the centers of these four circles to be 
fixed and that they turn in rolling contact; then if the point a on the 




circle A moves to a h a 2 , a 3 , the same point on B will move to 61, 6 2 , 63, 
and on C to Ci, C2, c 3 . Now if the point a on the circle C carries a marking- 
point, in its motion to d it will have traced from the circle A the hypo- 
cycloid aid, and at the same time from the circle B the epicycloid hci. 
This can be seen to be true if the circles A and B are now fixed; and if 
C rolls in A, the point Ci will roll to a h tracing the hypocycloid c\d\\ 
while if C rolls on B, ci will trace the epicycloid Cih. These two curves 



124 ' ELEMENTS OF MECHANISM 

in contact at d fulfil the fundamental law for tooth curves, that the 
normal to the two curves at the point Ci must pass through a. Simi- 
larly, if the original motion of the circles had been to a 2 , b 2 , c 2 , the same 
curves would be generated, only they would be longer and in contact 
at c 2 . If the hypocycloid c 2 a% is taken for the flank of a tooth on A, and 
the epicycloid c 2 6 2 for the face of a tooth on B, and if ctfH drives C2&2 
toward a, it is evident that these two curves by their sliding action, as 
they approach the line of centers, will give the same type of motion to 
the circles as the circles had in generating the curves, which was pure 
rolling contact. Therefore the two cycloidal curves rolled simultane- 
ously by the describing circle C will cause by their sliding contact the 
same angular speed ratio of A and B as would be obtained by A and B 
moving with pure rolling contact. 

If now the circles A, B, and D are rolled in the opposite direction to 
that taken for A, B, and C, and if the point a moves to a 4 , 64, and di on 
the respective circles, the point a on D while moving to d\ will trace 
from A the epicycloid audi, and from B the hypocycloid b^di. The curve 
adi may be the face of a tooth on A, and bS\ the flank of a tooth 
on B, the normal d x a to the two curves in contact at di passing through a. 
The flank and face for the teeth on A and B, respectively, which were 
previously found, have been added to the face and flank just found, 
giving the complete outlines, in contact at d\. 

If now the wheel B is turned L.H., the tooth shown on it will drive 
the tooth on A, giving a constant angular speed ratio between A and B 
until the face of the tooth on B has come to the end of its action with 
the flank which it is driving, at about the point c 2 . 

The following facts will be evident from the foregoing discussion: in 
the cycloidal system of gearing, the flank and face which are to act upon 
each other must be generated by the same describing circle, but the describ- 
ing circles for the face and flank of the teeth of one wheel need not be 
alike. The path of contact is always on the describing circles; in Fig. 149 
it is along the line diaci. 

139. Interchangeable Wheels. A set of wheels any two of which 
will gear together are called interchangeable wheels. For these the 
same describing circle must be used in generating all the faces and 
flanks. The size of the describing circle depends on the properties of 
the hypocycloid, which curve forms the flanks of the teeth (excepting 
in an annular wheel). If the diameter of the describing circle is half 
that of the pitch circle, the flanks will be radial (Fig. 150, A), which 
gives a comparatively weak tooth at the root. If the describing circle 
is made smaller, the hypocycloid curves away from the radius (Fig. 
150, B) and will give a strong form of tooth; but if the describing circle 



GEARS AND GEAR TEETH 125 

is larger, the hypocycloid will curve the other way, passing inside the 
radial lines (Fig. 150, C) and giving a still weaker form of tooth, and 
a form of tooth which may be impossible to shape with a milling- 
cutter. 

From the above the practical conclusion would appear to be that the 
diameter of the describing circle should not be more than one-half that 
of the pitch circle of the smallest wheel of the set. It will be found, 
however, that when the diameter of the describing circle is taken five- 
eighths the diameter of the pitch circle, the curvature of the flanks will 
not be so great, with the ordinary proportions of height to thickness of 
teeth, but that the spaces are narrowest at the bottom; this being the 
case, the teeth can be shaped by a milling-cutter 

In one set of wheels in common use the diameter of the describing 
circle is taken such that it will give radial flanks on a 15-tooth pinion, 




Fig. 150 

or five-eighths that of a 12-tooth pinion, the smallest wheel of the set. 
This describing circle has been used with excellent results. 

As an example, given an interchangeable set of cycloidal gears, 2-P., 
radial flanks on a 15-tooth pinion; a gear having 24 teeth is to drive 
one having 30 teeth. The diameter of a 2-P., 15-tooth pinion would be 
7| inches; to give radial flanks on this pinion the diameter of the de- 
scribing circle would be 3f inches. This is the diameter of the describ- 
ing circle for all the faces and flanks for any gear of the set. The 24-tooth 
gear will have a diameter of 12", and the 30-tooth gear will have 15" 
diameter. This will give the diagram in Fig. 151 ready for the rolling 
of the tooth outlines. 

140. To draw the Teeth for a Pair of Cycloidal Wheels, and to 
determine the Path of Contact- In Fig. 152 given the pitch circles 
A and B and the describing circles C and D, C to roll the faces for B 



126 



ELEMENTS OF MECHANISM 



and the flanks for A, while D is to roll the faces for A and the flanks 
for B. These curves may be rolled at any convenient place. In the 
figure, the wheel A is to be the driver and is to turn as shown. Choose 
any point, as b, on A and a point a on B at a distance from the pitch 
point af = bf. The epicycloid and hypocycloid rolled from a and b 
respectively, and shown in contact at b 2 , would be suitable for the faces 
of the teeth on B and the flanks of the teeth on A respectively, and could 
be in action during approach. The curves may be rolled as indicated 
by the light lines. The method used to roll these curves is shown in 
Fig. 153, where the circle C is tracing a hypocycloid on A from the 

point o. Assume the circle C to start 
tangent to A at o and to roll as shown, 
drawing it in as many positions as may 
be desired to obtain a smooth curve, 
and these positions do not need to be 
equidistant; thus in the figure the 
center of C is at b, c, and d for the three 
positions used. Since the circle C rolls 
on A, the distance measured on A from 
o to a tangent point of C and A lis 
equal to the distance measured on C 
from that tangent point to the hypo- 
cycloid. The method of spacing off 
these equal arcs for the successive 
positions is indicated in Fig. 153. 

Returning to Fig. 152, the circle D 
is to roll the faces for the teeth on A 
and the flanks for the teeth on B. 
These curves may also be rolled from 
any convenient points, as c and d equi- 
distant from f. The face thus found from A may be traced and then 
transferred to the flank already found for the teeth on A at the point b, 
giving the curve b 2 bc f , the entire acting side of a tooth on A. Similarly 
by transferring the flank dd 3 to the point a we have b 2 ad', the shape of 
the teeth for the wheel B. It will be seen that the face on A could have 
been rolled from b as well as from c, so that the entire tooth curve could 
be rolled from 6, and similarly the other tooth curve could have been 
rolled from the point a. After finding the tooth curves, and knowing 
the addendum, clearance, and backlash, the teeth may be drawn. In 
Fig. 152 the teeth are drawn without backlash, and in contact on their 
acting surfaces at h and k. The path of contact is efg on the describing 
circles and is limited by the addendum circles. 




Fig. 151 



GEARS AND GEAR TEETH 



127 



141. Limits of the Path of Contact. Possibility of Any Desired 
Action. If, in Fig. 152, the teeth of either wheel are made longer, the 
path of contact and arc of action are increased; the extreme limit of 
the path of contact would therefore be when the teeth become pointed. 

It is often desirable to find whether a desired arc of action in approach 
or in recess may be obtained before rolling the tooth curves. Given 
the pinion A driving the rack B as shown in Fig. 154; to determine if 




an arc of approach equal to ab is possible. The path of contact must 
then begin at c, where the arc ac is equal to ab. The face of the rack's 
tooth must be long enough to reach from b to c, and this depends on the 
thickness of the tooth measured on the pitch line, since the non-acting 
side of the tooth must not cause the tooth to become pointed before 
the point c is reached. To see if this is possible without the tooth curves, 
draw a fine from c parallel to the line of centers (in general this line is 
drawn to the center of the wheel; the rack's center being at infinity gives 



128 



ELEMENTS OF MECHANISM 



the line parallel to the line of centers), and note the point d where this 
line crosses the pitch line of the rack. If bd were just equal to one-half 
the thickness of the tooth, the tooth would be pointed at c, and the 
desired arc of approach would be just possible; if bd were less than one- 
half the thickness of the tooth, the tooth would not become pointed 
until some point beyond c was reached, so that the action would be pos- 
sible and the teeth not pointed, as shown by the figure. 

If it is desired to have the arc of recess equal to the arc af, then the 
path of contact must go to g, and the face of the pinion must remain in 
contact with the flank of the rack until that point is reached, or the 
face must be long enough to reach from / to g. Drawing a line from g 
to the center of the pinion A, we find that the distance fh is greater 




Fig. 153 

than one-half of fk, which is taken as the thickness of the tooth; there- 
fore the desired arc of recess is not possible even with pointed teeth. 

The minimum arc of action, as in the case of involute gears, is the 
circular pitch. 

142. Annular Wheels. Fig. 155 shows a pinion A driving an annu- 
lar wheel B, the describing circle C generating the flanks of A and the 
faces of B, which in an annular wheel he inside the pitch circle, while D 
generates the faces of A and the flanks of B. The describing circle C 
is called the interior describing circle, and D is called the exterior describ- 
ing circle. The method of rolling the tooth curves, and the action of 



GEARS AND GEAR TEETH 



129 



the teeth, are the same as in the case of two external wheels, the path of 
contact being in this case efg when the pinion turns R.H. If these 
wheels were of an interchangeable set, the describing circles would be 
alike and found as explained in § 139, and the annular would then gear 




Fig. 154 



with any wheel of the set excepting for a limitation which is discussed 
in the following paragraph. 

143. Limitation in the Use of an Annular Wheel of the Cycloidal 
System. Referring to Fig. 155, it will be evident that, if the pinion 
drives, the faces of the pinion and annular will tend to be rather near each 
other during recess (during approach also on the non-acting side of the 



130 ELEMENTS OF MECHANISM 

teeth). The usual conditions are such that the faces do not touch; but 
the conditions may be such that the faces will touch each other without 
interference, for a certain arc of recess; or, finally, the conditions may 
be such that the faces would interfere, which would make the action of 
the wheels impossible. 

To determine whether a given case is possible it is necessary to refer 
to the double generation of the epicycloid and of the hypocycloid. 
The acting face of the pinion, Fig. 155, is rolled by the exterior describ- 
ing circle D, while the acting face of the annular is generated by the 
interior describing circle C. Two such faces are shown in Fig. 156 as 
they would appear if rolled from the points g and h, equidistant from 
the pitch point k. The acting face of A is an epicycloid, and is made 
by rolling the circle D to the right on A. It is also true that a circle 
whose diameter is equal to the sum of the diameters of A and D would 
roll the same epicycloid if rolled in the same direction. This circle is 
E, Fig. 156, and is called the intermediate describing circle of the pinion. 
The acting face of the annular is a hypocycloid rolled by the interior 
describing circle C rolling to the left inside of B. It is also true that 
the same hypocycloid would be rolled by a circle whose diameter is 
equal to the difference between the diameters of B and C, provided it is 
rolled in the opposite direction. This circle is F, Fig. 156, and is called 
the intermediate describing circle of the annular. 

If now the four circles A, B, E, and F turn in rolling contact, through 
arcs each equal to kg, the point k will be found at g, h, m, and n on the 
respective circles, the point k on E having rolled the epicycloid gm, 
while k on F rolls the hypocycloid hn. 

To determine whether these faces do or do not touch or conflict, 
assume that the given conditions gave the circles E and F coincident 
as in Fig. 157 where 

diam. A + diam. D = diam. E = diam. B - diam. C = diam. F. 

Here if the three circles A, B, and (EF) turn in rolling contact, the 
point k moving to g on A will move to h on B and to (ran) on the com- 
mon intermediate circle. This means that the common intermediate 
circle could simultaneously generate the two faces; therefore the two 
faces are in perfect contact on the intermediate circle. This contact 
will continue until the addendum circle of one of the wheels crosses 
the intermediate circle, the addendum circle crossing first necessarily 
limiting the path of contact. 

The above may be stated as follows: If the intermediate describing 
circles of the pinion and annular coincide, the faces will be in contact in 
recess, if the pinion drives, in addition to the regular path of contact. 



GEARS AND GEAR TEETH 



131 




Fig. 155 



132 



ELEMENTS OF MECHANISM 




GEARS AND GEAR TEETH 



133 



If in Fig. 157 the exterior describing circle, for example, should be 
made smaller, as in Fig. 158, then the intermediate of the pinion would 
be smaller than that of the annular; but if the exterior describing circle 
is smaller, the face gm will have a greater curvature and will evidently 
curve away from the face hn, so that no contact between the faces can 
occur, as is shown in Fig. 158. Here no additional path of contact 




occurs, and it is evident, if the arcs kg, km, kn, and kh are equal as they 
must be, if the circles move in rolling contact, that the smaller D be- 
comes (and consequently E) the greater will be the space between the 



This may be stated as follows : If the intermediate describing circle of 
the pinion is smaller than that of the annular, the faces do not touch, and 
the action is in all respects similar to the cases of external wheels. 



134 



ELEMENTS OF MECHANISM 



In Fig. 159 the exterior describing circle D is made larger than it is 
in Fig. 157, so that the intermediate E of the pinion is larger than F, 
that of the annular. Making the circle D larger would give the face of 

1 




Fig. 159 

the pmion less curvature, which would cause tne curve gm to cross the 
curve hn, giving an impossible case. Therefore, if the intermediate of 
the pinion is greater than that of the annular, the action is impossible. 



GEARS AND GEAR TEETH 



135 



144. Low-numbered Pinions, Cycloidal System. The obliquity of 
action in cycloidal gears is constantly varying; it diminishes during the 
approach, becoming zero at the pitch point, and then increases during 
the recess. For wheels doing heavy work it has been found by experi- 
ence that the maximum obliquity should not in general exceed 30°, giving 
a mean of 15°. When more than one pair of teeth are in contact, a high 
maximum is less objectionable. 

As the number of teeth in a wheel decreases, they necessarily become 
longer to secure the proper path of contact, and both the obliquity of 




action and the sliding increase. From the preceding considerations the 
practical rule is deduced that, for millwork and general machinery, no 
pinion of less than twelve teeth should be used if it is possible to avoid it. 

It often becomes necessary, however, to use wheels having less than 
twelve teeth, in light-running mechanism, such as clockwork. In such 
cases a greater obliquity may be admissible, and for light work the flank- 
describing circle may be made large. 

Let it be required to determine the possibility of using two equal 
pinions, having six teeth, with radial flanks, the arcs of approach and 



136 



ELEMENTS OF MECHANISM 



recess each equal to one-half the pitch, and to find the maximum angle 
of obliquity. Fig. 160 is the diagram for two such gears. The path of 
contact is to begin at a, the arcs ab, cb, and db each being equal to one- 
half the pitch; then the face of the pinion B must be long enough to be 
in contact with the flank of A at a. Drawing the fine aef from a to the 
center of B, the distance de will be found to be less than one-half the 
thickness of the tooth. Therefore the approach is possible. Since the 
pinions are alike, the recess is also possible. The maximum angle of 




Fig. 161 

obliquity in approach is the angle 6, and this may be found in degrees 
as follows. The arc be on the pitch circle A subtends an angle bgc equal 
to one-half the pitch angle, the arc of approach being equal to one-half 
the pitch; in this case the angle bgc is 30°. The arc ab on the describ- 
ing circle C is equal to be and therefore subtends an angle bha, which is 
to the angle bgc inversely as the radii of the respective circles. In this 
case these radii are as 2 to 1, making the angle bha equal to 60°. (It is 
important to notice that the line gc does not pass through the point a 
excepting in the single case of a radial flank gear.) The angle 9 between 
the tangent and the chord ab will always be one-half the angle ahb sub- 
tended by the arc ab. This gives the angle of obliquity 30°. Therefore 
we find that two pinions with six teeth and radial flanks will work with 



GEARS AND GEAR TEETH 



137 



arcs of approach and recess each equal to one-half the pitch and with 
the maximum angle of obliquity of 30°. By allowing a greater angle of 
obliquity the teeth may be made a little longer and so give an arc of 
action greater than the pitch, which should be the case in practice. 

Two pinions with five teeth each will work with describing circles 
having diameters three-fifths the diameter of the pitch circles, and arcs 
of approach and recess each equal to one-half the pitch, as shown by 




Fig. 162 

Fig. 161, the path of contact beginning at a, the arcs ab, cb, and db each 
being equal to one-half the pitch. The action is possible, since de is 
less than one-half the thickness of the tooth. The maximum angle of 
obliquity is 30°, the angle bgc being 36° and bha being f of 36°, or 60°. 
Two pinions with four teeth each will just barely work with describ- 
ing circles having diameters five-eighths the diameter of the pitch circle, 
and with no backlash, the arcs of approach and recess each being one- 
half the pitch. Fig. 162 shows the diagram for this case, and the teeth 



138 



ELEMENTS OF MECHANISM 



are apparently pointed, which would be the case if de were just one-half 
the thickness of the tooth. To determine the possibility of the action 
the angle dfe may be calculated. It should not be greater than 22 J° 
to allow the desired arc of approach. It will be found to be 22° 27' 19", 
so that the action is just possible. The maximum angle of obliquity 
6 will be found to be 36°. 

A pinion with four teeth will work with a pinion having four teeth 
or any higher number, if the arc of action is not required to be greater 
than the pitch, the maximum angle of obliquity not exceeding 36°. 

The requirements may be very different from the above in every 
respect; an arc of action greater than the pitch would usually be re- 
quired; it might be desired to have the arc of recess greater than the 
arc of approach; it might not be admissible to have so great an angle 
of obliquity or to have the teeth cut under so far as a describing circle 
five-eighths the pitch circle would require. The results would of course 
vary with the conditions imposed. 

145. Stepped Wheels. If a pair of spur wheels are cut transversely 
into a number of plates, and each plate is rotated through an angle, 
equal to the pitch angle divided by the number of plates, ahead of the 
adjacent plate, as shown in Fig. 163, the result will be a pair of stepped 
wheels. This device has t] ie effect of increasing the number of teeth 

without diminishing their strength; and 
the number of contact points is also in- 
creased. The upper figure shows a section 
on the pitch line A A. The action for 
each pair of plates is the same as that for 
spur wheels having the same outlines. In 
practice there is a limit to the reduction 
in the thickness of the plates, depending 
on the material of the teeth and the pres- 
sure to be transmitted, since too thin plates would abrade. The number 
of divisions is not often taken more than two or three, and the teeth 
are thus quite broad. These wheels give a very smooth and quiet action. 

146. Twisted Spur Gears. If, instead of cutting the gear into a 
few plates, as shown in Fig. 163, the number of sections is infinite, the 
result is a helical gear such as that shown in Fig. 116. 

The twisting being uniform, the elements of the teeth become helices, 
all having the same lead, see §§ 161 and 162. The line of contact 
between two teeth will have a helical form, but will not be a true helix; 
the projection of this helix on a plane perpendicular to the axis will be 
the ordinary path of contact. It can easily be seen that the common 
normal at any point of contact can in no case lie in the plane of rota- 



r~ 


i i.l kJ: 


H 


/ m 


\m 


\ \c l 


3 \ 




\* — p-*- 


T> 





163 



GEARS AND GEAR TEETH 



139 




tion, but will make an angle with it. The line of action then can in 
general have three components: 1° A component producing rotation, 
perpendicular to the plane of the axes; 2° A component of side pressure, 
parallel to the line of centers; 3° A component of end pressure parallel 
to the axes. The end pressure may be neutralized as explained in 
§147. 

147. Herring-bone Gears. A gear like that shown in Fig. 117, 
known as a herring-bone gear, is equivalent to two helical gears, one 
having a right-hand helix and the other a left-hand helix. The use of 
a pair of gears of this type eliminates the end 
thrust on the shaft referred to in the preceding 
paragraph. 

148. Sliding Friction Eliminated. In Fig. 
164, which represents a transverse section of a 
pair of twisted wheels, suppose the original tooth 
outlines to have been those shown dotted. Then 
cut away the faces as shown by full lines having 
the new faces tangent to the old ones at the 
pitch point c; proper contact is lost except that 
at c for the section shown, but by twisting the wheels this contact can 
be made to travel along the common element of the pitch cylinders 
through c from one side of the wheel to the other. A simple con- 
struction to use in this case is to make the flanks of the wheels radial 
and the faces semicircles tangent to the flanks. The action here is 

purely rolling and is very smooth and noiseless; 
but for heavy work it is best to use the common 
forms of teeth with sliding action, so that the 
pressure may be distributed over a line instead 
of acting at a point. 

149. Pin Gearing. In this form of gearing 
the teeth of one wheel consist of cylindrical 
pins, and those of the other of surfaces parallel 
to cycloidal surfaces, from which they are 
derived. 

In Fig. 165 let o x and o 2 be the centers of 
the pitch circles whose circumferences are di- 
vided into equal parts, as ce and eg. Now if 
we suppose the wheels to turn on their axes, 
and to be in rolling contact at c, the point e of the wheel 0\ will trace 
the epicycloid gp on the plane of the wheel o 2 , and merely a point e 
upon the plane of o t . Let cf be a curve similar to ge and imagine a pin 
of no sensible diameter — a rigid material line — to be fixed at c in the 




Fig. 165 



140 



ELEMENTS OF MECHANISM 




upper wheel. Then, if the lower one turn to the right, it will drive the 
pin before it with a constant velocity ratio, the action ending at e if 
the driving curve be terminated at / as shown. 

If the pins be made of a sensible diameter, the outlines of the teeth 
upon the other wheel are curves parallel to the original epicycloids, as 
shown in Fig. 166. The diameter of the pins is usually made about 
equal to the thickness of the tooth, the radius being, therefore, about \ 
the pitch arc. This rule is, however, not 
imperative, as the pins are often made con- 
siderably smaller. 

Clearance for the pin is provided by forming 
the root of the tooth with a semicircle of a 
radius equal to that of the pin, the center being 
inside of the pitch circle an amount equal to 
the clearance required. 

The pins are ordinarily supported at each 
end, two discs being fixed upon the shaft for 
the purpose, thus making what is called a 
lantern wheel or pinion. 

In wheel work of this kind the action is 
almost wholly confined to one side of the fine 
of centers. In the elementary form (Fig. 165) 
the action is wholly on one side, and receding, since it cannot begin until 
the pin reaches c (if o 2 drives), and ceases at e; if Oi is considered the 
driver, action begins at e, ends at c, and is wholly approaching. As 
approaching action is injurious, pin gearing is not adapted for use where 
the same wheel has both to drive and to follow; the pins are therefore 
always given to the follower, and the teeth to the driver. 

When the pin has a sensible diameter, the tooth is shortened and 
its thickness is decreased; the line of action is also shortened at e, Fig. 
166, and, instead of beginning at c, will begin at a point where the 
normal to the original tooth curve, through the 
center of the pin, first comes in contact with 
the derived curve mf. This normal's end will 
not fall at c, but at a point on the arc ce beyond, 
on account of a property of the curve parallel to 
an epicycloid. The parallel to the epicycloid is 
shown in Fig. 167, cp being the given epicy- 
cloid. The curve may be found by drawing 
a series of arcs ss with a radius equal to the normal distance between 
the curves, and with the centers on cp. The parallel curve first 
passes below the pitch curve cm and then rises, after forming a cusp, 



Fig. 166 




Fig. 167 



GEARS AND GEAR TEETH 



141 



and cuts away the first part drawn: this is more clearly shown 
somewhat exaggerated at mno. Hence the part which would act on 
the pin when its center is at c is cut away, and, for the same epicycloid, 
the greater the diameter of the pin the more this cutting away. In 
Fig. 166 the pin e is just quitting contact with the tooth at i while c is 
at the pitch point, and, according to the above property of the parallel 
to the epicycloid, is not yet in contact with the tooth m. Strictly speak- 
ing, then, the case shown is not a desirable one, as the tooth should 
not cease contact at i until m begins its action. The above error is 
practically so small that it has been disregarded, especially for rough 
work. 

The following method may be used in determining a limiting case 
in pin gearing: 

If the pitch arc = eg is assumed (Fig. 168), the greatest possible 
height of tooth is determined by the intersection of the front and back 
of the tooth at p; and if this height 
is taken, action will begin at c and 
end at h, the point in the upper 
pitch circle through which p passes. 
Now if p falls upon the pitch circle 
ceh, we should have a limiting case 
for a pin of no sensible diameter. 
If the pin has a sensible diameter 
and the pitch arc eg = ce is assigned, 
bisect eg with the line o<ip and draw 
ce intersecting o 2 p in k; assume a 
radius for the pin less than ek and 
draw the derived curve to cut o%p 
in j, which will be the point of the 
tooth. Through j draw a normal 
to the epicycloid, cutting it at s; 
through s describe an arc about Oi 
cutting the upper pitch circle at t, 
the position of the center of the pin at 
the end of its action. Draw the outline mf of the next working tooth, 
find the point m at the cusp of the curve parallel to the epicycloid, and 
draw the normal mn; m is the lowest possible working point of the 
tooth. Through n describe an arc about o 2 cutting the original path 
of contact in r, which is the point that n must reach before the tooth will 
be in contact with the pin, or is the point that n must reach before the 
common normal to the pin and tooth curve passes through the pitch 
point. 




142 ELEMENTS OF MECHANISM 

Now action begins when the axis of the pin is at r and ends at t; if 
rt = ce, we have an exact limiting case and the assumed radius of the 
pin is a maximum; if rt < ce, the radius is too great; but if rt > ce, 
the case is practical To get the exact limit a process of trial and error 
should be resorted to. When the pin is a point the methods used in 
cycloidal gearing may be applied; the correction for a pin of sensible 
diameter can then be made by applying the method of Fig. 168. 

150. Pin Gearing: Wheel and Rack. As the pins are always given 
to the follower, two cases arise. 

1° Rack drives, giving the pin-wheel and rack, Fig. 169. Here the 
original tooth is bounded by cycloids generated by the pitch circle of 
the wheel. 

2° Wheel drives, giving the pin-rack and wheel. Here (Fig. 170) the 
original tooth outline is the involute of the wheel's pitch circle. 

151. Inside Pin Gearing. Here also there are two cases. 

1° Pinion drives (Fig. 171). The original tooth outlines will be inter- 
nal epicycloids generated by rolling the pitch circle of the annular 
wheel on the. pinion's pitch circle. 

2° Annular wheel drives (Fig. 172). Here the original tooth outline is 
the hypocycloid traced by rolling the pinion's pitch circle in the wheel's 
circle. 

Path of Contact. In the elementary form of tooth (Fig. 165) the path 
of contact is on the circumference of the pitch circle of the follower Oi, 
as ce. When a pin is used its center always lies in this circumference, 
and its point of contact may be found by laying off a distance ei equal 
to the radius of the pin (Fig. 173) on the common normal. Drawing a 
number of these common normals, all of which must pass through the 
pitch point c, and laying off the radius of the pin ei on each, we have 
the path of contact ci known as the limagon. 

152. Double-point Gearing. This form of gearing, shown in Fig. 
174, gives very smooth action where not much force is to be trans- 
mitted. The pitch circles are here taken as the describing circles; the 
face eg of the pinion Oi is generated by rolling the pitch circle o 2 on that 
of Oi, and the face cf is generated by rolling the pitch circle 0\ on o 2 . 
If Oi is considered the driver, action begins at d, the point c of Oi sliding 
down the face cf while c travels from d to c. In the receding action the 
point c of the tooth of o 2 is acted on by the face eg while c moves from 
c to e. The spaces must be so made as to clear the teeth. This com- 
bination reduces friction to a minimum and gives the obliquity of action 
less than in any case except pin gearing, but the teeth are much under- 
cut and weakened by the clearing curves, and if much force is to be 
transmitted the line of contact will soon be worn away. 



GEARS AND GEAR TEETH 



143 





Fig. 169 



Fig. 170 





Fig. 171 



Fig. 172 





Fig. 174 



144 



ELEMENTS OF MECHANISM 



153. Bevel Gears. A pair of bevel gears bears the same relation 
to a pair of rolling cones that a pair of spur gears bears to rolling cylin- 
ders. 

Fig. 175 shows two bevel gears meshing together, the contact in this 
case being internal. Here, as in the case of spur gears, the angular 
speeds are inversely proportional to the number of teeth. 

The pitch circle of a bevel gear is the base of the cone which the gear 
replaces. 

154. To Draw the Blanks for a Pair of Bevel Gears. A convenient 
way to gain an understanding of the principle of bevel gear design will 

be to study the method of 
drawing the blanks from 
which a pair of bevel gears is 
to be cut. Let it be assumed 
that a 6-pitch, 12-tooth gear 
is to mesh with an 18-tooth 
gear, the axes to intersect at 
90°. Start with the point 0, 
Fig. 176, as the point of inter- 
section of the two axes. Draw 
OS and OSi making the re- 
quired angle (in this case 90°). 
These are the center lines of 
the shafts. Assume that the 
12-tooth gear is to be on Si. 
Call this gear A and the 18- 
tooth gear B. Since A has 
12 teeth and is 6-pitch, its 
pitch diameter, that is, the 
diameter of the base of its 
pitch cone, is 12 4- 6 or 2 in. 
In like manner the pitch 
diameter of B is 18-^-6 or 
3 in. From measure along 
OSi a distance OM equal to the pitch radius of B (1| in.) and, through 
M , thus found, draw a line perpendicular to OSi. In like manner make 
ON equal to the pitch radius of A and draw a line through N perpen- 
dicular to OS. These lines intersect at K. Make MR equal to MK and 
NT equal to NK. From R, K, and T draw lines to Q. Then the tri- 
angle ORK is the projection of the " pitch cone" of the gear A and OTK 
that of the "pitch cone" of B. It will be noticed that the above con- 
struction is the equivalent of that for rolling cones. Next, draw through 




Fig. 175 



GEARS AND GEAR TEETH 



145 



K a line perpendicular to OK meeting OSx at P and OS at H (not shown) . 
Draw a line from H through T and from P through R. The cone rep- 
resented by the triangle THK is called the normal cone of the gear B 
and that represented by the triangle RPK is called the normal cone of A. 
These cones will be explained more in detail later. 

From R lay off Ra equal to the addendum that is to be used on gear 
A (this is determined by the same considerations that would be used 




Fig. 176 

for the addendum on a spur gear). Along RO lay off Rr equal to the 
desired length of gear face (§113). Through r draw a line parallel to 
PR. From a draw a line to meeting this parallel at a x . Through a 
draw a line parallel to RK meeting PKH at a 2 . From ax draw a line 
parallel to RK meeting a line drawn from a 2 to at a 3 . Lay off along 
RP the distance Rd equal to the dedendum and draw from d toward 
meeting air at dx. Find d 2 and d 3 in the same way that a 2 and a 3 were 



146 



ELEMENTS OF MECHANISM 



found. The figure add&i represents the tooth. The dimensions of the 
hub and the position of lines FG and FiQi and of the corresponding 
lines on the other gear may be made anything that is desirable. 

It will appear from this construction that bevel gears must be laid 
out in pairs. 

155. Teeth for Bevel Gears. The gear blanks, as laid out in § 154, 
are of the form ordinarily used, and the information there given is per- 
haps all that a person making use of bevel gears would need. In order 
to understand the principles underlying the action of the gears it may 
be desirable to notice the relation between bevel gear teeth and spur 
gear teeth. 

In the discussion on the teeth of spur wheels, the motions were con- 
sidered as taking place in the plane of the paper, and lines instead of 
surfaces have been dealt with. But the pitch and describing curves, 
and also the tooth outlines, are but traces of surfaces acting in straight- 
line contact, and having their elements perpendicular to the plane of 
the paper. In bevel gearing the pitch surfaces are cones, and the teeth 
are in contact along straight lines, but these lines are perpendicular to 
a spherical surface, and all of them pass through the center of the 
sphere, which is at the point of intersection of the axes of the two pitch 
cones. 

In Fig. 177, is the center of the sphere, AOC and BOC are the pitch 
cones. If the teeth are involute, cones such as MON and KOL are the 
base cones, and the teeth may be thought 
of as being generated by a plane rolling 
on each of the base cones, the ends of the 
teeth lying on the surface of the sphere, 
and the tooth outlines being the curves 
traced on this surface by the plane which 
generates the teeth. 

156. Drawing the Teeth on Bevel 
Gears. Tredgold's Approximation. Since 
narrow zones of the sphere, Fig. 177, 
near the circles BC and AC, will nearly 
coincide with cones whose elements are 
tangent at B, C and M, the conical sur- 
faces may be substituted for the spherical ones without serious error, 
and as the tooth outlines are always comparatively short they may 
be supposed to lie on the cones. These cones BPC and AHC are called 
the normal cones and correspond to the cones RPK and THK of Fig. 
176. Fig. 178 shows the method of drawing the tooth outlines. It will 
be noticed that the developments of portions of the normal cones are 




GEARS AND GEAR TEETH 



147 



treated as if they were pitch circles of spur gears and the teeth are 
drawn on the development exactly as if they were teeth of spur gears, 
and are then transferred to the other views by ordinary principles of 
projection. 

157. Crown Gears. When the angle at the apex of the cone of one 
of a pair of bevel gears is 180° the pitch cone becomes a flat disk and 
the normal cone becomes a cylinder. Such a gear is analagous to a 
rack bent in the form of a circle. The teeth taper inward, elements of 




Fig. 178 



the teeth converging toward the center of the disk. Another bevel of 
any number of teeth may be designed to run with a crown gear but the 
angle between the axes will depend upon the ratio of the teeth. Fig. 
179 shows such a pair of gears. 

158. Twisted Bevel Gears. The teeth of bevel gears may be 
twisted in the same manner as the teeth of spur gears (see § 146). Fig. 
122 shows a pair of twisted bevels such as used for the drive to the 
differential of an automobile. The gears from which the drawing was 
made were from a Pierce Arrow touring car. 

159. Skew Bevels. Fig. 123 shows a pair of skew bevel gears used 
in cotton machinery. Here the shafts are at right angles, non-inter- 



148 



ELEMENTS OF MECHANISM 



secting, but passing so near each other that ordinary helical gears can- 
not be used. 

The pitch surfaces of these gears are hyperboloids of revolution, and 
the teeth are in contact along straight lines. The angular speeds are 
inversely as the pitch diameters. 

160. Screw Gearing. This class of gearing is used to connect non- 
parallel and non-intersecting shafts and includes the two types known 
as worm and wheel and helical gears. The latter when used for this 
purpose are often, but inaccurately, called spiral gears. In the helical 

gears and the elementary forms 
of worm and wheel the teeth 
have point contact. The speed 
ratio is not necessarily in the 
inverse ratio of the diameters. 
The action of gears of this class 
is similar to the action of a 
screw and nut which will be 
considered in a later chapter. 
This is particularly evident in 
the case of the worm and 
wheel. The distinction be- 
tween the worm and wheel and 
the helical gears, however, is 
not a very clear one, being 
largely a matter of speed ratio 
and manner of forming the 
teeth. Both may properly be 
considered here as helical 
gears and the following dis- 
FlG - 179 cussion will apply to both 

The worm and wheel will be considered in Chapter VIII as a screw 

and nut. 

161. The Helix; Its Construction and Properties. A helix is a 
curve wound around the outside of a cylinder or cone advancing uni- 
formly along the axis as it winds around. The nature of the curve 
and the method of drawing it may be understood from a study of 
Fig. 180. 

The angle of a helix is the angle which a straight line tangent to the 
helix at any point makes with an element of the cylinder. This angle 
is the same for all points on a cylindrical helix. Two helices are said 
to be normal to each other when their tangents drawn at the point 
where the helices intersect are perpendicular to each other. 




GEARS AND GEAR TEETH 



149 



When two parallel lines wind around a cylinder forming parallel 
helices, as in Fig. 181, the result would be called a double helix; three 
lines would give a triple helix, and so on. If the helix slopes as in Fig. 




Fig. 180 

180 it is called a right-hand helix; if it slopes in the reverse direction 
it is called a left-hand helix. 

162. Lead. Axial Pitch. The distance L, Figs. 180 and 181, by 
which a helix advances along the axis of the cylinder for one turn 
around is called the lead. The distance A, measured, parallel to the 
axis, from one point on a helix to the corresponding point on the 
next turn of a single helix, or to the corresponding point on the next 
helix in the case of a multiple helix, is called 
the axial pitch. In the case of a single helix 
this is equal to the lead; in the case of a 
double helix the axial pitch is equal to one- 
half the lead; in a triple helix, one-third 
of the lead, and so on. 

163. Normal Pitch. The distance P, 
between a point on a helix to the corre- 
sponding point on the next turn of a single 
helix or the corresponding point on the 
next helix in the case of a multiple helix, measured along the normal to 
the helix, is called the normal pitch. 

164. Helical Gears are gears whose teeth wind partially around the 
pitch cylinders. A pair of such gears may be used to connect parallel 
shafts, as shown in Fig. 116, or non-parallel shafts, which is the case 
now under discussion. The method of forming the teeth and the 
action of the teeth differ in the two cases.* The definitions given 

* The twisted gears (Fig. 116) have line contact between teeth while the helical 
gears in general have point contact, or multiple point contact. 




150 



ELEMENTS OF MECHANISM 



above apply to the teeth of helical gears. In order that two helical 
gears may work together they must have the same normal pitch and 
the angle between the shafts must be such that the tangent to the pitch 
helices of the two gears coincide at the pitch point. From this it follows 

Drbrea Skai 




that the sum of the angles of their helices must be equal to the angle 
between the shafts, or the supplement of this angle. 




Fig. 183 



Fig. 182 shows the pitch cylinders of a pair of helical gears. The 
line MN is the common tangent to the teeth at the point of contact of 



GEARS AND GEAR TEETH 151 

the pitch cylinders (that is, the pitch point). B is therefore the angle 
of the helix of the driver and Bx the angle of the helix of the driven 
gear. Here the angle 6 between the two shafts is equal to 180° 
- (B + Bx). 

Fig. 183 is the development of the surfaces of the two pitch cylin- 
ders shown in Fig. 182, the slanting lines being the development of 
imaginary helices at the centers of the teeth on the pitch cylinders. 
The perpendicular distance between these lines is the normal pitch P 
(the same in both gears). The distance C and Cx between the points 
of intersection of two adjacent teeth with the ends of the cylinder 
(EF and ExFx) are the circular pitches of the respective gears. It will 
be noticed that the circular pitches of the two gears are not alike but 
depend upon the helix angles. 

It should be noticed that if the relation between the angles and /3 is 
such that ft becomes the driven gear becomes like an ordinary spur 
gear. Hence a properly formed worm may be made to drive a spur gear 
if their axes are set at the proper angle with each other. 

165. Relation between the Circular Pitches of a Pair of Helical 
Gears. Referring still to Fig. 183 



c = 


= 5 and C\ ■- 

cos B 


P 

cos Bx 


Cx 


cos B 




c 


cos Bx 





Therefore, -~ = r- . (52) 



166. Relation between Numbers of Teeth in a Pair of Helical 
Gears. Let T represent the number of teeth in the driver (Fig. 183) 
and Tx the number of teeth in the driven gear. 

Then 



Therefore 



T - 


= -jr an d Tx = 
■kDx 


irDx 
Cx 


Tx 
T ' 


Cx DxC 
tD DCx 

c 




C 


cos B\ 




Cx 


cos B 




Tx 


Dx cos Bx 




T 


D cos B 





But from equation (52) -^ = 

Therefore r-^Sf- (53) 

That is, the numbers of teeth are mmm& y as the product of the pitch diam- 
eters multiplied by the cosines of the helix angles 



152 ELEMENTS OF MECHANISM 

167. Speed Ratio of Helical Gears. As in the case of spur gears, 
the angular speeds of a pair of helical gears are inversely as the numbers 
of teeth. If N represents the angular speed of the driver, and Ni of the 
driven gear, it follows from equation (53) that 

N 1= D cosB 

N Dx cos B x w } 



CHAPTER VI 
WHEELS IN TRAINS 



168. Train of Wheels. A train of wheels is a series of rolling cylin- 
ders or cones, gears, pulleys or other similar devices serving to transmit 
power from one shaft to another. 

The examples of roiling cylinders, gears, etc., which have been dis- 
cussed in earlier chapters are really wheel trains each involving only 
one pair of wheels. In Fig. 184 D is a gear fast to shaft A. E is a gear 
fast to shaft B and meshing with D. F is another gear also fast to shaft 
B and meshing with the gear G 
which is fast to shaft C. If now 
the shaft A begins to turn, D will 
turn with it, and, therefore, cause 
E to turn. Since E is fast to the 
shaft B the latter will turn with E. 
Gear F will then turn at the same 
angular speed as E and will cause G 
to turn, causing the shaft C to turn 
with it. That is, D drives E, and F, 
turning with E, drives G. 

The above is an example of a 
simple train of gears consisting of 
two pairs. Fig. 185 shows an 
arrangement of pulleys similar in 
action to the gears shown in Fig. 
184. H is a pulley on the shaft R 
belted to the pulley J on shaft S. On the same shaft is another pulley 
K belted to the pulley L on shaft T. 

Fig. 186 shows a train of wheels involving both gears and pulleys. 
In this case D is a gear on shaft A, meshing with and driving the gear 
E on shaft B. On the same shaft is pulley K, belted to the pulley L on 
shaft C. 

169. Driving Wheel and Driven Wheel. Referring again to Fig. 
184, the gear D by its rotation causes E to turn; therefore, D may be 
called the driver or driving wheel, and E the driven wheel. Similarly F, 
turning with E, is the driver for the wheel G. Hence, in any train such 

153 




Fig. 184 



154 



ELEMENTS OF MECHANISM 



as here shown, consisting of three axes with two pairs of wheels, two of 
the wheels are drivers and two are driven wheels. 

170. Idle Wheel. In Fig. 187, gear D drives E, which in turn 
drives F. E is, therefore, both a driven and a driving wheel. Such 
a wheel is called an idle wheel. 

171. Train Value (Speed Ratio). The ratio of the angular speed 
of the last wheel of a train to the angular speed of the first wheel of the 




Fig. 185 

same train is called the value of the train, or train value, and will be rep- 
resented by e. For example, if the shaft A in Fig. 184 makes 25 r.p.m., 
and the sizes of the several gears are such that shaft C makes 150 r.p.m., 




Fig. 186 

the value of the train would be -^ = 6 = e. An inspection of the 
same figure will show that if A turns right handed, B will turn left 
handed and C will turn right handed. The direction, then, of C is the 
same as that of A. The value of this train is then said to be positive, 




WHEELS IN TRAINS 155 

and will be indicated by putting a + sign in front of the number which 
indicates its value. If the number of wheels involved is such that the 
last shaft turns in the opposite direction from the first shaft, the value 
of the train will be said to be negative, which fact will be indicated by 
a — sign in front of the number indicating the train value. 

172. Calculation of Speeds. Let 
it be assumed that the gears in 
Fig. 184 have teeth as follows: /^ ^\ ^-4--^/£ 

D, 100 teeth F, 125 teeth 

E, 50 teeth G, 25 teeth 

It will also be assumed that shaft 
A makes 25 r.p.m. and it is re- 
quired to find the speed of C. ' F 1S7 
Since the speed of B is to the 

speed of A as the teeth in D are to the teeth in E, the revolutions of B 
will be equal to 25 X -V°o°-J a l so > since the speed of C is to the speed of E 
as the teeth in F are to the teeth in G, the speed of C = 25 X -V\r- 
X W- = 250. Expressing this as a formula, 

The speed of the last shaft is equal to the speed of the first shaft 

the product of the teeth of all the drivers . . 

the product of the teeth of all the driven wheels 

In the case of pulleys in Fig. 185 the principle is the same, except 
that diameters are used instead of numbers of teeth. Suppose that 
pulley H is 24 ins. in diameter, J 8 ins. in diameter, K 36 ins. in diameter, 
and L 12 ins. in diameter, then the speed of T will be equal to the speed 

of R X -5 zr^r ; that is, in the case of a train of pulleys: 

The speed of the last shaft is equal to the speed of the first shaft 

the product of the diameter of all the driving pulleys , . 

the product of the diameter of all the driven pulleys 

In a train consisting of a combination of gears and pulleys, as in Fig. 
186, 

The speed of the last shaft is equal to the speed of the first shaft 
the product of diameters and numbers of teeth of all the driving wheels , . 
the product of diameters and numbers of teeth of the driven wheels 

An idle wheel such as gear E in Fig. 187 has no effect on the speed 
ratio, but does cause a change in the direction. This can be seen from 
the following calculation: Let the wheel D have 100 teeth; E 75 teeth, 



156 



ELEMENTS OF MECHANISM 




and F 25 teeth, then the speed of shaft C is equal to the speed 
of A X V/ X H- 75 tnen > which is the number of teeth in the idle 
wheel, appears in both numerator and denominator and cancels out, 

and therefore the speed of C be- 
comes the speed of A X ^J-. 

173. Driving and Driven Gears 
having Coincident Axes. Fig. 188 
is a diagram of the back gear 
arrangement for a simple cone 
pulley head-stock on an engine 
lathe. It illustrates the principles 
involved when two wheels, whose 
axes coincide, are connected by a 
train of wheels through an inter- 
mediate shaft, the axis of the 
intermediate shaft being parallel 
to the axis of the connected wheels. P is the cone pulley, A is a 
gear integral with P and meshing with gear B. C is another gear on 
the same shaft with B, both B and C being fast to the shaft. C meshes 
with gear D on the spindle S. From Eq. 55. 
Speed of spindle = speed of cone pulley 

Teeth in A X Teeth in C 
Teeth in B X Teeth in D 
Since, however, the shaft R is parallel to S the gears must be so pro- 
portioned that the pitch radius of A + pitch radius of B equals pitch 
radius of C + pitch radius of D. Consequently, if the pitches of the 
two pairs are to be in some definite ratio there must be a corresponding 
relation between the sum of the teeth in A and B and the sum of the 
teeth in C and D. 

174. Frequency of Contact between Teeth: Hunting Cog. Given 
two gears Gi and G 2 with teeth 7\ and T 2 having the greatest common 
divisor a. 

Then let Ti = ah and T% = ak. 

Therefore Turns Gl = Tl = a Jl = k 

inereiore, Turns (? 2 T x at, k 

Contact between the same pair of teeth will take place after the passage of 
a number of teeth equal to the least common multiple of 7\ and T 2 = atit 2 . 
Therefore, before the same pair of teeth come in contact again after 
having been in contact, 

atit 2 
ati 
ahh 
at 2 



the turns of Gi 



and turns of G 2 



= U 



WHEELS IN TRAINS 157 

Therefore the smaller the numbers h and t 2 , which express the velocity 
ratio of the two axes, the more frequently will the same pair of teeth be 
in contact. 

Assume the velocity of ratio of two axes to be nearly as 5 to 2. Now, 
if T 7 ! = 80 and T, = 32, 

T, 80 5 .. 

Y 2 = 32 = 2 GXaCtly ' 
and the same pair of teeth will be in contact after five turns of T 2 or 
two turns of 7\. 

T^ 81 5 
If Ti is changed to 81, then =- = ^r = -> very nearly, the angular 

1 2 oJj Z 

velocity ratio being scarcely distinguishable, from what it was originally. 
But now the same teeth will be in contact only after 81 turns of T 2 or 
32 turns of TV 

The insertion of a tooth in this manner prevents contact between the 
same pair of teeth too often, and insures greater regularity in the wear 
of the wheels. The tooth inserted is called a hunting cog, because a 
pair of teeth, after being once in contact, gradually separate and then 
approach by one tooth in each turn, and thus appear to hunt each other 
as they go round. In cast gears, which will be more or less imperfect, 
it would be much better if an imperfection on any tooth could distribute 
its effect over many teeth rather than that all the wear due to such 
imperfection should come always upon the same tooth. This result is 
most completely obtained when the numbers of teeth on the two gears 
are prime to each other, as above, when Ti and T 2 were 81 and 32 
respectively. 

175. Example of Wheels in Trains. The following paragraphs will 
give a few examples of wheels in trains. These are selected because 
they serve to illustrate the principles involved, and not because a knowl- 
edge of these particular trains is of special importance. 

176. Clockwork. A familiar example of the employment of wheels 
in trains is seen in clockwork. Fig. 189 represents the trains of a com- 
mon clock; the numbers near the different wheels denote the number 
of teeth on the wheels near which they are placed. 

The verge or anchor vibrates with the pendulum P, and if the 
pendulum vibrates once per second, it will let one tooth of the escape- 
wheel pass for every double vibration, or every two seconds. Thus the 
shaft A will revolve once per minute, and is suited to carry the second 
hand S. 

1"iirns (j 
The value of the train between the axes A and C is T 

8X8 1 ns 

= fif> pA = ™> or the shaft C revolves once for sixty revolutions of 



158 



ELEMENTS OF MECHANISM 



D 



A ; it is therefore suited to carry the minute hand M. The hour hand 
H is also placed on this shaft C, but is attached to the loose wheel F by- 
means of a hollow hub. This wheel is connected to the shaft C by- 
means of a train and intermediate shaft E. The value of this train is 
turns H _ 28 X 8 _ J_ 
turns M ~ 42 X 64 ~ 12 ' 

The drum D, on which the weight-cord is wound, makes one revolu- 
tion for every twelve of the minute hand M, and thus revolves twice 
each day. Then, if the clock is to run 
eight days, the drum must be large enough 
for sixteen coils of the cord. The drum 
is connected to the wheel G by means of 
a ratchet and click, so that the cord can 
be wound upon the drum without turning 
the wheel. 

Clock trains are usually arranged as 
shown in the figure, the wheels being placed 
on shafts, often called "arbors," whose 
bearings are arranged in two parallel plates 
which are kept the proper distance apart 
by shouldered pillars (not shown) placed 
at the corners of the plates. When the 
arbor E is placed outside, as shown, a 
separate bearing is provided for its outer 
end. 

177. Planer Drive. Fig. 190 shows a 
portion of the gear train for driving the 
platen on a planer. The shaft S is driven 
by a motor or by a counter-shaft above 
the machine. S drives Si by the pair of 
gears A and B. On Si are two gears E 
and D which are fast to each other and 
slide on a long key on the shaft so that 
they are obliged to turn with the shaft, but 
may be moved along it by means of a 
shifting device fitting into the groove V. 
A similarly arranged pair of sliding gears 
H and K are on the shaft S 2 . Gears B 
and C are permanently attached to Si and 
F and G to S 2 . The shaft S 2 has a pulley on the end (not shown in 
the diagram) which drives, by a belt, to the mechanism operating the 
table. The object of the system of gears shown in the diagram is to 



•6 



Fig. 189 



WHEELS IN TRAINS 



159 



enable the operator to obtain four different speeds for $2, and therefore 
for the platen, for one speed of S. 

With H and K in the position shown and with D and E to the left so 
that E meshes with F as shown, E is driving F and the train value 

between S and $ 2 is 

Teeth A X Teeth ff 
Teeth B X Teeth F 

All the other gears are turning idle. If D and E are moved to the right 
a distance just a little more than the width of the face of E this gear 



Driving Shaft 



Fig. 190 

will be out of mesh with F and the whole system will be in neutral posi- 
tion. If D and E are moved still further to the right D will mesh with 

A D 

G and the train from S to $ 2 is ~ X ~ • 

Now, putting D and E back into neutral, slide H and K to the right 

A B 
until K meshes with B and the train between S and S 2 is -5 X -^- If 

# and 2£ are slid to the left H will mesh with C and the train will be 

B X H 

The levers which operate the two, sliding units are interlocked in 
such a way that one of the units must be in neutral position before the 
other can be moved to the right or left of neutral. 



160 



ELEMENTS OF MECHANISM 



178. Automobile Transmission. Fig. 191 is a diagram of the ar- 
rangement of the gears in a common type of automobile transmission 
which allows three speeds forward and one reverse. 

The gear A is on the end of a sleeve driven directly from the motor. 
A turns freely on the end of the propeller shaft P. On the counter- 
shaft S are four gears, B, D, F, and G which are attached to each other 
and turn as a unit. The purpose of the system of gears is to make it 
possible for A, turning at a definite speed, to drive shaft P at three 




Fig. 191 

different speeds in the same direction that A itself is turning, and at 
one speed in the direction opposite that of A. 

B is in mesh with A and the counter-shaft unit is turning in a direc- 
tion opposite that of A at a speed which is to the speed of A equal to 
Teeth in A 

Teeth in B ' gears in positions shown, the counter- 

shaft unit is turning idle and the whole system is in neutral. On the 
left end of A is the hub M which has teeth on its circumference. In 



WHEELS IN TRAINS 



161 



the right end of gear C is a cavity with spaces into winch the teeth of 
M can lock if C is moved to the right. C slides on a key, or keys, in shaft 
P so that when C turns P turns with it. If the gear-shifting lever is 
moved to the proper position C will be pushed over M thus locking C 
to A and the shaft P will be driven at the same speed as A. This puts 
the transmission in "high." If C is pushed to the left of its present 
position it will mesh with D and then 

Speed P _ Teeth in A X Teeth in D 
Speed A ~ Teeth in B X Teeth in C ' 



-[ 



A 2>" 



This gives the "second" or "intermediate" speed. If C is put in the 
position shown and the shifting lever moved so as to take hold of gear 
E and move it to the right to mesh with F 
the speed ratio will be 

Speed P = Teeth in A X Teeth in F 
Speed A Teeth in B X Teeth in E ' 

This gives the "third" or "low" speed. 
Pushing E to the left of its neutral posi- 
tion brings it into mesh with the idler K 
which is driven from G giving a speed 
ratio 

Speed P _ Teeth in A X Teeth in G 
Speed A ~ Teeth in B X Teeth in E ' 



^2ZI 



B 4" 



and since the connection between G and 
E is through the idler, the direction of 
E and therefore of P is opposite that 
of A, giving the " reverse " drive. 

179. Cotton Card Train. Fig. 192 shows the train in a cotton card- 
ing-machine. For the train we have the value 



Fig. 192 



turns B 
turns A 



^x^x 
17 X 20 X 



130 17 



26 



X 



33 



+37.84. 



In the machine the lap of cotton passing under the roll A is much drawn 
out in its passage through the machine, and it becomes necessary to 
solve for the ratio of the surface speeds of the rolls B and A . For this, 
since the surface speed equals 2 j X turns X radius or turns X t X 
diameter, 

surface speed B _ turns B X diam. B 

surface speed A 



surface speed B _ 4 

surface speed A 2.25 



turns A X diam. A 
67.27. 



162 



ELEMENTS OF MECHANISM 



180. Hoisting Machine Train. A train of spur gears is often used 
in machines for hoisting where the problem would be to find the ratio 
of the weight lifted to the force applied. In Fig. 193 the value of the 
train is 



41 



turns B 



21 25 _1_ 

100 X 84 ~ 



2pt then, if 



turns A 100 " 84 16 ' 
D = 15" and R = 1| ft., 
speed W = 1_ 15 = J_. 
speed F 16 30 32' 

F_ = speed W = l 
" W speed F 32' 

or if F were 50 lbs., W would be 1600 lbs. 

if loss due to friction were neglected. 
181. Reduction Gear. Fig. 194, furnished by the Fellows Gear 
Shaper Co., shows a train of gears consisting of a pinion concentric with 
an annular, with three idle gears. The gears in this case are herring- 
bone gears. • 



Fig 



W 
193 




182. Designing Gear Trains. No definite rules or formulas can be 
followed in designing a train of gears to have a certain train value. The 
process is mainly one of " cut and try" until the desired result is obtained. 
There are, however, some general lines of attack which may be followed, 



WHEELS IN TRAINS 



163 



and which may best be understood by studying certain typical prob- 
lems. If the value of the train is chosen arbitrarily, it may be found 
impossible to select gears which will give exactly the value called for. 



Example 24. Let it be required to select the gears for a train in which the last 
gear shall turn 19 times while the first gear turns once, the direction of rotation being 
the same. No gear is to have less than 12 teeth nor more than 60 teeth. 

Solution. The first step in 
the solution of this problem is 
the determination of the num- 
ber of pairs of gears necessary 
to give a train value of 19 and 
keep the gears within the limits 
of size specified. If only one 
pair were used, making the 
driver as large as allowed, that 
is, 60 teeth, and the driven 
gear as small as allowed, that 
is, 12 teeth, the train value 
would be f§ or 5. If a second 
driver of 60 teeth is made fast 

to the 12 teeth gear, and this drives a second gear of 12 teeth, as shown in Fig. 195, 
the train value becomes ft X ft = 25. This is greater than the assigned value of 
19, therefore, two pairs of gears will be sufficient to Obtain a value of 19 without 
exceeding the specified limits of size. Having thus determined the number of gears 
necessary, the next step is the selection of the gears themselves to give the exact 
value of 19. This may be tried in several ways. First, since two pairs of gears are 

to be used, the square root of 19 
may be taken. This is approxi- 
mately 4.36. This does not differ 
greatly from 4f. Now 
,19 





f^t 



19. 



Multiply both numerator and 
denominator of the first fraction 
by 12, and of the second fraction 
by 3 and the equation becomes 

Fig. 196 « X « = 19- 

Therefore, a train of gears as 
shown in Fig. 196 will give the required train value of 19. If the directional relation 
were not correct an idler might be used. 

Example 25. Let it be required to find suitable gears to drive a shaft at 4 r.p.m. 
from a shaft making 500 r.p.m., using no gear of less than 24 teeth nor more than 96 
teeth. 

Solution. The train value in this case is ^fo = tib- 

Since the smallest gear that can be used is 24 teeth and the largest 96 teeth, the 
train value between any gear and its driver cannot be less than f| or \. 



164 



ELEMENTS OF MECHANISM 



Therefore £ must be multiplied by itself a sufficient number of times to obtain 
a final result less than T f s . Now, (|) 3 = ^; then 3 pairs of gears will not be enough. 
But (|) 4 =2^6 j which is beyond the specified T |g, hence 4 pairs will be proper to use. 

1 3 

The fourth root of j^-g is approximately ^ or — • 

A train of gears for a trial might be in the ratio of 

lvlviyiL = ivlvivi 
10 10 10 (A) 3 10 10 A 10 27' 

Multiplying numerator and denominator of each of the first three terms by 9 and of 
the last term by 3 gives. 

87 V » V 2! V '* 
90 A 90 A 9 J A ay. 

Then a train, such as shown in Fig. 150, fulfils the required conditions. 




Fig. 197 



Example 26. Where an error of a certain amount is allowable, as would very 
often be the case, the following method may be used to advantage. 



T 100 

Let the value of the train be 60 and — = -^r 



5, 



where T represents the maximum number of teeth and t the minimum. It will be 
found that three pairs of gears are needed. Therefore take the cube root of 60, 
which is 3.91 +, and write 



3.91 N , 3.91 vx 3.91 
1 X 1 X 1 



), nearly. 



Since the small gears are not to have less than 20 teeth, and since 20 X 3.91 = 78+, 
a first approximation may be written 

iixiix n, 

which will be found to equal 61.63; if this result is too greatly in error, a reduction 
of one or two teeth in the numerator or an increase in the denominator may give a 
closer result, as 

11 x n x n = 60.07. 



WHEELS IN TRAINS 165 

Example 27. To design a train of four gears, with the axis of the last wheel coin- 
cident with the axis of the first wheel as in Fig. 188. The train value to be T V No 
gear to have less than 12 teeth. All gears to be of the same pitch. 

Solution. Since there are two pairs in the train, the value fa must be separated 
into two factors and it is desirable to have these factors as nearly as may be of the 
same value. The square root of fa is between f and f so a trial pair of factors may 
be taken |X|. Then letting the letters T a , T b , T c , Ta represent the numbers of 
teeth in the gears A, B, C, D respectively (Fig. 188). 
Ta^Tc 11 

Now, since the pitches are all alike, 

T a + T b = T c + Ta (see § 173). 

Let Z represent this sum. Then a value must be chosen for Z such that it may be 

broken up into two parts whose ratio is \ and also two parts whose ratio is \. 

If Z is made equal to the least common multiple of 1 + 3 and 1+4, the condi- 

T a 5 T c 4 1 

tion will be satisfied. This L. C. M . is 20. Then -~r would be — and ;=- would be j^ • 

But these values are too small for the numbers of teeth in the gears. Then 
numerator and denominator of both fractions must be multiplied by some number 
such that no number will be less than the number of teeth allowed in the smallest 
gear. In this case multiplying T 5 5 and T 4 6 each by f gives \% and \\. 

Therefore T a may be 15, T b = 45, T c = 12, Ta = 48. 

The above method (Example 27) may be expressed as follows: 
When both pairs have the same pitch. 

If 7 X 7 are the factors of e (i.e., the train value) expressed in lowest 

terms, then T a + T b and T c + T& must be made equal to the L. C. M. of 
ti + 1 2 and h + U or to some multiple of the L. C. M. 

When the pitches of the two pairs are different. 

The case illustrated in Example 27 is not a practical one, because the 
stresses on the second pair of gears are always greater, requiring a 
greater circular pitch. 

If the pitch number of A and B = Pi and the pitch number of C and 

D = P 2 , and if — ■ = -^ reduced to its lowest terms, then T a + T b is 

Pi -T2 

made equal to the L. C. M. of t\ + 1 2 and U + U (or to some multiple 
of the L. C. M) multiplied by p h and T c + T d is made equal to the 
L. C. M. (or the same multiple of the L. C. M.) multiplied by p%- 



CHAPTER VII 



EPICYCLIC GEAR TRAINS 

183. An epicyclic train of gears is a train in which some of the 
gears turn on fixed axes, while others turn on axes which are them- 
selves in motion. The wheels are usually connected by a rigid link 
known as the train arm which rotates on the axis of one of the 
wheels of the train. 

Assume that C, Fig. 198, is a gear carried by the arm A and pinned 
to A so that it cannot turn on its own axis. If A is caused to turn 
about S once, a reference mark V on C, which in the position shown is 
pointing downward, would point in every direction successively as A 

revolved and come finally to its 
present position when the arm had 
made a complete turn. 

If C is meshing with another gear 
on the axis S, and therefore turns 
on its axis at the same time that the 
axis revolves, the reference mark 
will swing around and point to its 
original direction a number of times 
equal to the algebraic sum of ,the 
speed of the arm and the speed of 
C on its axis. This resultant number of times that the reference mark 
returns to its original direction, having always turned completely 
over, is called the number of absolute turns that C makes, or its absolute 
speed. The speed of C on its own axis is called its relative speed, or speed 
relative to the arm. Either direction of rotation may be assumed as 
positive (+) ; then rotation in the opposite direction must be considered 
as negative ( — ). 

Fig. 199 illustrates an epicyclic train, and the following description 
of its operation should be studied carefully in order to understand the 
principle of action. When this principle is clear the analysis of any 
epicyclic train is comparatively simple. 

B is a gear turning with the shaft S which is in stationary bearings 
and is driven by the gears R and K. C is a gear meshing with B. The 
stud T on which C turns is carried by the arm A. Fast to the hub of 
A is the gear E, driven by the gear D. Attached to C is the gear F 
which drives G. G turns on the same axis as B, but must, of course, 
be free to turn at a different speed from B. 

166 




EPICYCLIC GEAR TRAINS 



167 



D and R receive their motion from outside sources. The resultant 
speed of G, due to the combined speeds of B and A, is the algebraic 
sum of the speeds which it would have when each moved with the other 
standing still. 

If the gear D is first assumed not to turn, the arm A will be station- 
ary, and the following equation will hold true 

Speed of G Teeth in B X Teeth in F 



Speed of B Teeth in C X Teeth in G 
Speed of G = speed of B X train value. 



(I) 



ti 




Fig. 199 
If, on the other hand, the gear B is assumed not to turn and D turns 
at a definite speed, the arm A will revolve, the stud T will travel around 
S as an axis, C rolling around on B, and F rolling around on G. This 
will impart motion to G which may be found as follows : Suppose A to 
have a speed of a r.p.m. right-handed; then the speed of C on its own 
axis is the same as if A were held still and B turned with a r.p.m. left- 
handed. That is, C would have a speed on its axis of a X ™ — -r~ ■ — Ti 

v Teeth in C 

r.p.m. right-handed, relative to its own axis, or relative to the arm. 
This speed of C would impart to G, relative to the arm, a speed of 

a X ^ xU . — ^ w m — nr^ — T< r.p.m. = a X Train value, left-handed. 
Teeth in C X Teeth inG r 



168 ELEMENTS OF MECHANISM 

But the arm is itself turning right-handed at a speed a r.p.m. Therefore, 
the actual speed of G due to the speed of A is 

a —a X train value. (II) 

Combining (I) and (II) 

Speed of G = Speed of B X Train Value + Speed of Arm 

— Speed of Arm X Train Value. (58) 

If n represents the absolute turns or speed of the last wheel of an 
epicyclic train (in this case G), m the absolute turns or speed of the 
first wheel (B), a the turns or speed of the arm (A), and e the train 
value, Equation (58) may be expressed thus 

n = me + a — ae. (59) 

Equation (59) is commonly written 

. = ^ (60) 

m — a 

and, in this form, may be expressed in words thus: 

„ . . _ Turns last wheel relative to arm 
Turns first wheel relative to arm 
_ Absolute turns of last wheel — turns of arm 
Absolute turns of first wheel — turns of arm 

Problems relating to epicyclic trains may be solved by the formula 
given in Equation (59) or (60) or by another method known as the 
tabulation method. This consists in assuming that the motions of 
B and the arm (Fig. 199) take place successively instead of simul- 
taneously. The gears are first assumed to be made fast to the arm so 
that there can be no relative motion. The arm is made to turn at the 
proper speed for a unit of time in the proper direction; all the gears 
will turn with it. Then the arm is held still and one of the gears (in 
this case B) is turned backward or forward enough to make its net 
number of turns equal to its known speed. The sum of the results 
produced by these two processes gives the net motion or speed of the 
other gear or gears. 

If m, n, a, and e have the same meanings as in Equations (59) and 
(60) the above process may be tabulated thus: 

Turns of Arm Turns of B Turns of G 

1° Train locked a a a 

2° Arm fixed __0 m — a ^ (m — a) e 

3° Resultant motions a m 1 (m — a) e + a 



^jlA^M^-X^ 



(XA^kiv- 



EPICYCLIC GEAR TRAINS 169 

If the resultant number of turns of G thus obtained be equated to n 
the resulting equation is 

n = (m — a) e + a 

n — a 



which is the same as Equation (60). 

It is absolutely essential that the + or — sign precedes each of the 
quantities according as its value is positive or negative. 

In applying either the formula or the tabulation method care must 
be taken to include only those gears which are a part of the epicyclic 
train. For instance, in Fig. 199 the gears D, E, R and K serve merely 
as drivers of members of the epicyclic train and do not enter into the 
formula or the tabulation. 

184. Solution of Problems on Epicyclic Trains. The following ex- 
amples will illustrate the application of the two methods of solving 
epicyclic trains. In some cases the formula is used and in other cases 
the tabulation method, while in a few examples both methods are used. 
Either method will apply to any problem and it is often desirable to 
solve by both for the purpose of checking. 

Example 28. In Fig. 199 let B have 80 teeth, C 40 teeth, F 90 teeth and G 30 
teeth. If B has a speed of 100 r.p.m. right-handed and A 60 r.p.m. left-handed, 
let it be required to find the speed of G. 

Solution No. 1. Using Eq. (59) let right-handed rotation be assumed plus. Then 
m = +100, a = — 60, e = f£ X |§ = 6, and since with the arm at rest G would 
turn in the same direction as B, the value of e is plus. That is, e = +6. 
Substituting in Eq. (59), 

n = 100 X 6 + (-60) - (-60 X 6) = 600 - 60 + 360 = 900. 

Solution No. 2. Using the tabulation method. 

Arm B G 

Gears locked to arm — 60 — 60 — 60 

Arm held still +160 160 X 6 



- 60 + 100 960 - 60 
= 900 

Example 29. In Fig. 200 let gear B have 24 teeth and C 18 teeth. If B is held 
from turning and the arm makes 1 turn right-handed, let it be required to find how 
many absolute turns C makes. 

Solution No. 1. Using Eq. 59, m = 0, e = — f f = — f, a = +1 (assuming 
right-hand rotation is plus). Then, substituting in Eq. (59), n = + 1 — (— f), 
or, n = |. 

Solution No. 2. Using the tabular method: 

Arm B C 

Gears locked to arm +1 +1 +1 

Arm held still ^1 - 1 (- |) 

+ 1 +1 



170 



ELEMENTS OF MECHANISM 




Fig. 200 



Example 30. In Fig. 201 E is an annular gear which cannot turn, being fast to 
the frame of the machine. The arm A turns about the shaft S which is also the 
axis of the gears B and E. B has 24 teeth, C 20 teeth, D 16 teeth, and E 96 teeth. 
Let it be required to find the speed of the arm A to cause the gear B to have a speed 
of 75 r.p.m. left-handed. 

Solution. Assume B to be the first wheel of the train and assume right-handed 

rotation as + : 

Then referring to Eq. (59) n = 0, m = 
-75, e = +|| = +i. 

Substituting these values in the 
equation, 

0- 75 Xl + a-aXh 
whence 

a = +25. 

Therefore, A will have to have a speed of 
25 r.p.m. right-handed to give the required 
speed to B. 

Example 31. Sun and Planet Wheel. Fig. 202 shows an applica- 
tion of the two-wheel epicyclic train known as the Sun and Planet 
Wheels, first devised by James Watt to avoid the use of a crank, 
which was patented. In his device the epicyclic train arm was replaced 
by the stationary groove G, which kept the two wheels in gear, a 
represents the engine shaft, to 
which the gear D was made fast, 
B the connecting rod, attached 
to the walking beam. The gear 
C was rigidly attached to the end 
of the connecting rod. While 
with such an arrangement it is 
not strictly true that the gear C 
does not turn, yet its action on 
the gear D for the interval of one 
revolution of the epicyclic arm 
(that is, the line joining the 
centers of D and C) is the same 
as though C did not turn, since 
the position of C at the end of 
one revolution of .the arm is the 
same as at the beginning. 

Let it be assumed that the gears 
C and D have the same number of 
teeth. 

Then the train value = — 1. Let the arm ab make one turn. 

Required to find the turns of D and therefore of the engine shaft. 




EPICYCLIC GEAR TRAINS 



171 



Let m represent the turns of D, a the turns of the arm, n the turns of C. 
Solution. From equation (60), 

-1-JL4 

m — 1 
whence m = +2. 

That is, the engine shaft will make two turns every time the gear C passes around it. 
Example 32. In the three-wheeled train, Fig. 203, let A have 55 teeth, and C 
have 50. A does not turn. To find the turns of C while the arm D makes +10 
turns. 

Solution. Using the tabular method 

A C Arm 

Train locked +10 +10 +10 

Train unlocked, arm fixed —10 — 10 (ff) 

Resultant motions — 1 +10 

Or the wheel C turns —1 while the arm D turns +10. 

If the gear C in Fig. 203 were given the same number of teeth as A, 
it would not turn at all. If there were more teeth in C than in A its 
resultant number of turns would be in the same direction as the arm. 




Fig. 202 



Fig. 204 



Example 33. Ferguson's Paradox. In the device shown in Fig. 
204, known as Ferguson's Paradox, all three of the cases just referred 
to occur in one mechanism. « 

Let the gear A have 60 teeth, C 61 teeth, E 60 teeth, F 59 teeth. B is an idle 
wheel connecting each of the others with A. The arm D turns freely on the axis 
of A and carries the axis which supports the other gears. A is fixed to the stand 
and therefore cannot turn. If the arm D is given one turn R.H. (+), required to 
find the turns of C, E, and F. 

Solution. Using the tabular method 



Train locked 

Train unlocked, arm fixed , 
Resultant motions 



A 

+ 1 
- 1 



D 

+ 1 




C 

+ 1 



E 

+ 1 
- 1 



F 

+ 1 







+ 1 







A 



172 



ELEMENTS OF MECHANISM 



Example 34. Ford Transmission. Fig. 205 is a diagram of the 
planetary or epicyclic transmission as used in the Ford automobile. 
The shaft P, fast to the engine shaft, carries the piece L on which are 
clutch rings. Between these rings are other rings which turn with 
the drum K. The latter is fast to the propellor shaft M. Powerful 
springs urge the clutch rings together, except when prevented by the 
levers W. The friction between the clutch rings serves to connect L to 

G 

P 33T. 

D Revers-e Low Gear Brake 

'ty ^'ii ^ m 




Fig. 205 

K, thus making a direct connection between the engine shaft and the 
propellor shaft, furnishing the direct drive to the latter, which in turn 
drives the rear wheels. This gives "full gear" forward. 

Fast to the front side of K is the sleeve S carrying the gear E, which 
meshes with the gear B running loosely on the stud H. This stud is 
carried by the piece A which is fast to the engine shaft. B is integral 
with the gears C and D. C meshes with F on the sleeve N fast to the 
drum J. D meshes with G on the hub of the drum I. A, turning with 
the engine and carrying the shaft H with it, constitutes the arm of two 
epicyclic trains. One of these trains is through the gears F, C, B, and 
E. The other train is through the gears G, D, B, E. 



EPICYCLIC GEAR TRAINS 173 

When the left pedal in the car is pressed forward a short distance 
the lever W is operated to release the pressure of the springs which 
hold the clutch rings in contact. The engine then runs idly without 
turning the propellor shaft. A further motion of the left pedal applies 
a brake band to the drum J, holding it from turning, thus holding the 
gear F at rest. C, rolling around the stationary gear F, and B around 

E, gives a speed to E dependent upon the relative numbers of teeth in 

F, C, B, and E. 

This speed may be calculated as follows: 

Using Eq. (60) m = turns of F = and n the turns of E. 

Assuming one turn of the engine, a = +1, e = +f| X f f = + tt- 

7 n — 1 4 ■ 4 

Then — r = _ 1 whence n = +rr* That is, the gear E makes ry of a turn 

for every turn of the engine, in the same direction as the engine, and carries the 
propellor shaft with it. 

If the left pedal is allowed to come back to its normal position, re- 
leasing the drum J, and the hand lever operated to hold out the clutch, 
then if the middle pedal is pressed forward, it applies a brake to the drum 
I, thus holding the gear G stationary. This brings into action the 
epicyclic train G, D, B, E, giving motion to E in a direction opposite 
that of the engine. 

This may be calculated as follows: 

Letting to represent the turns of G and using the same formula as before 

t" — ~zi A 27 — $• 

5 n - 1 1 

v = -x z ' whence n = — - • 

4 0-1 4 

That is E, with the propellor shaft, makes | turn for each turn of the engine and in 
the opposite direction. 

When the right pedal is pressed forward a brake is applied to the 
drum K, retarding or stopping the propellor shaft. When this is done 
the drives should of course all be in neutral. 

Example 35. All Spur Gear Differential. Fig. 206 is a diagram of 
the arrangement of gears in an automobile differential where no bevel 
gears are employed except the pair which transmits the motion from 
the propellor shaft. The large bevel gear A (usually a twisted bevel) 
is driven from the propellor shaft and carries the cage B which always 
turns with A. This cage supports the studs S on which are the small 
gears C and D. There are several pairs of these small gears equally 
spaced around the cage but one pair will be sufficient to consider in 
discussing the action of the mechanism. C meshes with the gear E 



174 



ELEMENTS OF MECHANISM 



which is on the axle L, C also meshes with D, as shown, and D in turn 
meshes with the gear F on the axle R. 

When the car is moving in a straight path so that the rear wheels 
are turning at the same angular speed, the cage B and all the gears 
inside it revolve as a unit and there is no relative motion of the gears. 
When the car starts to turn a curve, the wheel which is on the outside 
of the curve must travel farther than the inner wheel, and therefore 



Bearing 




Fig. 206 

must make more turns in a given time. The gears E, C, D, and F then 
begin to turn relative to each other and to the cage and the action 
becomes that of an epicyclic train with the cage, carrying the studs S, 
as the epicyclic arm. 

This action can be seen more clearly if the car is supposed to be standing still with 
the right wheel (on the axle R) jacked up clear of the ground while the left wheel 
rests on the ground and is blocked to prevent it from rolling. Then E may be con- 
sidered as the first wheel of the epicyclic train, C and D intermediate wheels, and F 
the last wheel. The train value between E and F is — 1 since the gears E and F 
have the same number of teeth. Then, using equation (60) where m = the turns 
of gear E, a = turns of the cage and n the turns of F. 

m =0 , a = +1 (assuming the large bevel to make one turn) . 

n — \ 
Whence — 1 = - or n = +2. 



Therefore with one wheel at rest the other wheel will turn twice as fast as the 
large bevel. 



EPICYCLIC GEAR TRAINS 



175 



With both wheels turning, but one of them turning faster than the other, similar 
relative motion takes place. For example, suppose that the car is turning a corner 
such that the right wheel must turn twice as fast as the left one. 
Then, using the same notation as above, 

n = 2m', 
2m — a 



whence 



1 = 



m — a 
f a. 



Example 36. Triplex Pulley Block. Fig. 207 shows a vertical sec- 
tion and side view, with part of the casing removed, of a triplex pulley 




Fig. 207 



block. S is the shaft to which the hand chain wheel A is keyed. Also 
keyed to S is the gear F meshing with the two gears E. The gears E 
turn on studs T which are carried by the arm B, the latter being keyed 
to the hub of the load chain wheel G. The gears C are integral with E 
and mesh with the annular D which is a part of the stationary casing. 
The mechanism is an epicyclic train. F is the first wheel of the train 
and has a speed imparted to it by the turning of the hand chain wheel 
A. The annular D is the last wheel of the train and does not turn. 
The train value is 



Teeth in F Teeth in C 
'Teeth in E X Teeth in D ' 



176 



ELEMENTS OF MECHANISM 




Assuming one turn of A, the turns of the arm B may be found, and, 
therefore, the turns of G. Hence, knowing the angular speed of A and 
its diameter, and the angular speed of G and its diameter, the relative 
linear speeds of the hand chain and the load chain can be calculated. 
The load will then be to the force exerted on the hand chain as the 
speed of the hand chain is to the speed of the load chain, friction being 
neglected. 

185. Epicyclic Bevel Trains. Fig. 208 represents a common form 
of epicyclic bevel train, consisting of the two bevel-wheels D and E 
attached to sleeves free to turn about the shaft extending through 

them. This shaft carries the cross 
at F which makes the bearings for 
the idlers GG connecting the bevels 
D and E (only one of these idlers 
is necessary, although the two are 
used to form a balanced pair, thus 
reducing friction and wear). The shaft F may be given any number 
of turns by means of the wheel A, at the same time the bevel D may 
be turned as desired, and the problem will be to determine the resulting 
motion of the bevel E. The shaft and cross F here correspond with 
the arm of the epicyclic spur-gear trains. 

When the bevels are arranged in this way the wheels D and E must 
have the same number of teeth, and the train value is —1. It will be 
found clearer in these prob- 
lems to assume that the 
motion is positive when the 
nearer side of the wheel 
moves in a given direction, 
say upward, in which case 
a downward motion would 
be negative; or if a down- 
ward motion is assumed as 
positive, then upward 
motion would be negative. 

Example 37. In Fig. 209 
B and E are two bevel gears 
running on shaft S, but not fast 
to it. Attached to the collar P, 
which is set screwed and keyed 
to S, is a stud T on which turns 

freely the gear D meshing with B and E. B and E are of the same size and T is at 
right angles with S. J is a gear having 25 teeth and driving the 40-tooth gear K 
which is fast to B. L is a 51-tooth gear driven by the 17-tooth gear H which is fast 




EPICYCLIC GEAR TRAINS 



177 



to E. AT is a 45-tooth gear fast to the same shaft as J and drives the 20-tooth gear 
M which is fast to S. It is required to find the speed of L if J makes 40 r.p.m. 

Solution. The first step is to pick out those gears which are a part of the epicyclic 
train. These are evidently B, D, and E. The epicyclic arm is T. Assume B as 
the first wheel of the epicyclic train, E the last wheel, and, letting m represent the 
speed of B, n the speed of E, a the speed of S and e the train value between E and 
B. Also assume direction in which J turns as positive. Using Eq. (59), 



m = —|f X 40 r.p.m. = —25 r.p.m., 
a = -M X 40 = -90. 

Then, substituting in Eq. (59), 

n = (-25) X(-l) +(-90) - {(-90) X(-l)} 
= 25-90-90 
= —155 r.p.m. = speed of E. 
Speed of L = -155 X (~W = 51f. 
Therefore, L has a speed of 51f r.p.m. in the same direction as J. 

This problem may be solved by the tabulation method also, the process being 
the same as for an all spur epicyclic train. 

Example 38. Bevel Gear Differential. Fig. 210 shows the arrange- 
ment of gears in the differential of an automobile. Shaft S is driven 
from the motor and has keyed to it the bevel gear D meshing with E 
which turns loosely on the 
hub of the gear H, the 
latter being keyed to the 
axle of the left wheel. 
E has projections on it 
which carry the studs T 
furnishing bearings for the 
gears R. There are several 
of these gears in order to 
distribute the load. The 
gears R mesh with H which 
is, as has been said, fast 
to the axle of the left 
wheel, and with K which is 

fast to the axle of the right wneel. When the automobile is going straight 
ahead D drives E and all the other gears revolve as a unit with E with- 
out any relative motion. As soon, however, as the car starts to turn 
a corner, say toward the right, the left wheel will have to travel further, 
and therefore the shaft B must turn faster than C. Then the gears 
begin to move relative to each other, the action being that of an epi- 
cyclic train. 



To Left Wheel 
B 




To Right Wheel 
■ 



Fig. 210 



178 



ELEMENTS OF MECHANISM 



Let it be assumed that the right wheel is jacked up so that the axle C and gear 
K may turn freely, while the left wheel remains on the ground and is held from 
turning, thus holding gear H from turning. Consider H as the first wheel of the 
train, E being the arm. Required to find the turns of C for one turn (+) of E. 

Solution. Using equation (60) 

-1 



n — l . 
- l' 

Whence n = 2. 

That is, the right wheel will turn twice as fast as the gear E. 

Example 39. Water Wheel Governor. An epicyclic bevel train 
has been used in connection with a train containing a pair of cone pulleys, 

in a form of water-wheel 
governor for regulating the 
supply of water to the 
wheel. Fig. 211 is a dia- 
gram for this train, the 
position of the belt con- 
necting the cone pulleys 
being regulated by a ball 
governor connecting by 
levers with the guiding 
forks of the belt. The 
governor is so regulated 
that when running at the 
mean speed the belt will be 




Fig. 211 



in its mid-position, at which place the turns of E and D should be equal, 
and opposite in direction, in which case the arm F will not be turning. 
If the belt moves up from its mid-position, and if A turns as shown, 
the arm F will turn in the same direction as the wheel E. 



With the numbers of teeth as shown in the figure, let it be required to find the 

ratio of the diameters - if C is to turn downward once for 25 turns of A in the direc- 
x 

tion shown; also to determine whether the belt shall be crossed or open. 

Solution. Let E be considered as the first wheel of the train. Then, to use 
equation (59), 

n = turns A X f ? = 25 X ff downward (+). 
e = —1, a = 1. 



turns;! X^X^ 
x 67 



25X-X^- 
Z5 X 67 X x 



Then, substituting in equation (59), 
30 



25 X 



67 



-•( 26x |xf) 

25 X f £ - 2 _ i 
25X1$ 



+ l-(-D 



375 



EPICYCLIC GEAR TRAINS 



179 



The minus sign in this value of - signifies that the value m (in which - first ap- 
pears) must be negative; that is, E must turn in the opposite direction from D. Hence 
the cone B must turn in the same direction as A and the belt be open. 

Example 40. The bevel train may be a compound train, as shown in Fig. 212. 
Here the'train value, instead of being —1, is — W 5 - X ft = ~~ " 6 v> if Z> is considered 
as the first wheel. 



42T. / 









\H\ V8T/ 
)\ \D 


15T. 




'kl 




F 





V 

Fig. 212 



Letting m represent the turns of E, n the turns of D, and a the turns of the arm 
(same as of C) and using Equation (60), and assuming A to make + 40 turns and B 
to make — 10 turns, 

_60 40 -a 
9 -10 - a ' 
Whence a =- -\% -. 

Or C will turn J/s - times in the same direction as B and E. 



CHAPTER VIII 



INCLINED PLANE, WEDGE, SCREW, WORM AND WHEEL 

186. Inclined Plane and Wedge. The inclined plane and wedge 
will be considered only as mechanical elements for producing motion 
or exerting force. In this sense they act essentially the same. In 
Fig. 213, P represents a wedge, or solid, whose lower surface mn is 
horizontal, resting on a horizontal surface XX and free to be moved 
along that surface. The upper surface mo is inclined at an angle with 
the horizontal. In Fig. 213 the back surface no is perpendicular to mn. 
S is a slide which may move up or down 
in the guides G, the lower end being 
inclined or beveled at the same angle as 
the upper surface of P, on which it rests. 
Suppose that P is moved to the left a 
distance mm h so as to occupy the position 
shown by the dotted lines. It is evident 
that S is forced up a distance ddi. If the 
length b and height a of P are known, it 
is possible to calculate the amount S will 
move for any known movement of P. 
Draw a vertical line mt meeting m x oi at t. Then mt = ddi since they 
are sides of a parallelogram. The triangles mmt and min x oi are 




Fig. 213 



evidently similar. 


Therefore, 






mt mmi 






Oin x mini 


But 




Oini = on, 
mt = ddi 


and 




mini = mn. 


Therefore 




ddi _ mmi 
on mn 


or 


ddi 


w on 

= mmi X — = mmi tan omn, 



(61) 



or, in words, the distance the slider rises is equal to the distance the wedge 
moves multiplied by the ratio of the height of the wedge to its length. 

180 



INCLINED PLANE, WEDGE, SCREW, WORM AND WHEEL 181 



In Fig. 214 a wedge is shown in which the end no is not perpendicular 
to mn. The same method of calculating the rise of the slider S would 
be used as in the previous case except that the vertical height ok is 
used in place of the length no, the shape of the back end of course 
having no effect on the motion of S. 

The wedge in Fig. 215 is itself raised when pushed to the left, due 
to its sliding upon the inclined stationary surface of K, and carries S 




Fig. 214 




up with it. It also gives an additional rise to S due to the slant of the 
surface mo. The resultant rise of S is, therefore, the sum of the two. 

It should be noticed that the above laws hold true only when the 
direction of motion of the slider S is perpendicular to the direction in 
which the wedge moves. 

187. Screw Threads. If the top surface mo of the wedge shown 
in Fig. 213 is assumed to be covered with a very thin strip of flexible 
material and this strip is wound around a cylinder whose circumference 
is equal to the length b, the angle of 
inclination with the horizontal remaining 
the same, it will assume a helical form as 
shown in Fig. 216. If the slide S has a 
point which reaches out and rests on the 
top surface of the strip, and the cylinder 
is turned in the direction of the arrow, S 
will be raised. The action of the helical 
surface on S is exactly the - same as the 
action of the wedge in Fig. 213. One 
complete turn of the cylinder will raise S 
a distance of P. One-half a turn will 




Fig. 216 



raise S a distance — > and so on 



In Fig. 217 a similar arrangement is 
shown, except that, in this case, S is stationary and the cylinder is free 



182 



ELEMENTS OF MECHANISM 




Fig. 217 



to move endwise as well as turn, the weight of the cylinder resting 
on the point of S through the helical blade. Now, if the cylinder 
is given one turn in the same direction as before, it will be lowered a 
distance P. 

A more exact description of the surface 
would be to say that it is generated by a 
radial line always perpendicular to the 
axis of the cylinder and with its inner 
end in contact with a helix of lead = P 
on the surface of the cylinder. 

Fig. 218 shows a cylinder with a strip 
wound around it in the same way as in 
the preceding figures only the strip here 
is very much thicker and is wound 
around several times. In actually making 
such a cylinder of metal a solid cylinder of diameter D would be taken 
and a helical groove cut around it, the metal left between the successive 
turns of the groove thus constituting the " helical strip." A cylinder 
so formed is called a screw, the projecting part, which we have called 
the helical strip,' 
being known as 
the screw thread. 
The action of 
such a thread on 
its follower is 
exactly the same 
as just described 
for Fig. 216 or 
217. Instead of 
using^a single pro- 
jecting point or 
surface for the 
thread to act 
against, as has 
been assumed in 
these figures, a hole with a corresponding thread inside it is formed, 
this thread being of the proper size, slant, and shape to just fit into 
the grooves of the screw. The piece which contains such a hole is 
known as a nut. (See Fig. 219.) 

188. Forms of Screw Threads. There are several forms of 
threads in general use. The more common ones are shown in 
Figs. 220 to 223. In Fig. 220 is shown the square thread used for 




Fig. 218 



Fig. 219 



INCLINED PLANE, WEDGE, SCREW, WORM AND WHEEL 183 

supporting or moving a load as in a jack-screw. In Fig. 221 is shown 
the thread ordinarily known as the Acme thread which is similar to 
the square thread except that its sides slope slightly, giving a stronger 
thread and making it possible to open and close a split nut around it. 
Such a thread is used on the lead screw of a lathe and in similar places 
where the screw moves the carriage and where it is necessary to sepa- 
rate the two halves of 
the nut on which it 
acts when it is desired 
to break the connec- 
tion between the screw 
and the carriage. 

Figs. 222 and 223 
show the V thread, 
which is the kind 
commonly used on 
bolts, machine screws, 
and, in fact, for most 
purposes where the screw and nut serve for holding purposes. It is 
also used in light apparatus for causing motion. These two forms are 
alike except for a slight difference in the angle of the sides and a 
difference in shape at the point and root. Figs. 224 to 227, with the 
accompanying tables, show the shapes and standard proportions of the 
above forms of threads. 

U. S. STANDARD SCREW THREADS 




Fig. 220 



Fig. 221 



Fig. 222 Fig. 223 




Fig. 224 



B = | V = f P nearly. 



Dia. 


Threads 


Dia. 


Threads 


Dia. 


Threads 


Dia. 


Threads 


Dia. 


Threads 


Screw. 


per In. 


Screw. 


per In. 


Screw. 


per In. 


Screw. 


per In. 


Screw. 


per In. 


i 


20 


3 


10 


1* 


6 . 


3 


31 


5 


21 


5 
16 


18 




10 


1* 


5* 


3i 


31 


5i 


2h 




16 


§ 


9 


If 


5 


3^ 


3| 


5* 


2| 


JL. 


14 


16 


9 


H 


5 


3f 


3 


5? 


2| 




13 


1 


8 


2 


4| 


4 


3 


6 


2| 


9 


12 


If 


7 


21 


4| 


4| 


2| 






5 


11 


li 


7 


2h 


4 


4| 


2| 






11 

16 


11 


11 


6 


2| 


4 


4f 


2f 







184 



ELEMENTS OF MECHANISM 



WHITWORTH OR ENGLISH STANDARD SCREW THREAD 

A = J = 0.16 P, 
6 

B = 0.64 P r = 0.137 P. 

i 

Fig. 225 




Dia. 


Threads 


Dia. 


Threads 


Dia. 


Threads 


Dia. 


Threads 


Dia. 


Threads 


Screw. 


per In. 


Screw. 


per In. 


Screw. 


per In. 


Screw. 


per In. 


Screw. 


per In. 


1 


20 


5 


11 




8 


if 


5 


3 


31 


5 


18 


A! 


11 


1| 


7 


1* 


4| 


3i 


3| 


3 


16 




10 


1| 


7 


2 


4| 


3§ 


3| 


A 


14 


H 


10 


1| 


6 


2i 


4 


3f 


3 


i 


12 


i 


9 


1? 


6 


n 


4 


4 


3 


9 
16 


12 


15. 
16 


9 


1 5 

1 8 


5 


n 


3i 







ACME THREADS 




Threads. 


B 


T 


K 


s 


N 


Threads. 


B 


T 


K 


's 


' N 


16 


0.048 


0.023 


0.018 


0.039 


0.044 


If 


0.323 


0.232 


0.226 


0.393 


0.399 


10 


0.060 


0.037 


0.032 


0.063 


0.068 


11 


0.343 


0.247 


0.242 


0.419 


0.425 


9 


0.066 


0.041 


0.036 


0.070 


0.075 


1t 5 t 


0.354 


0.255 


0.250 


0.433 


0.438 


8 


0.073 


0.046 


0.041 


0.079 


0.084 




0.385 


0.278 


0.273 


0.472 


0.477 


7 


0.081 


0.053 


0.048 


0.090 


0.095 


lf\ 


0.416 


0.301 


0.296 


0.511 


0.516 


6 


0.093 


0.062 


0.057 


0.105 


0.110 




0.448 


0.324 


0.319 


0.551 


0.556 


5h 


0.104 


0.070 


0.064 


0.118 


0.123 




0.479 


0.348 


0.342 


0.590 


0.595 


5 


0.110 


0.074 


0.070 


0.126 


0.131 




0.510 


0.371 


0.366 


0.629 


0.635 


4| 


0.121 


0.082 


0.077 


0.140 


0.145 


if 


0.541 


0.394 


0.389 


0.669 


0.674 


4 


0.135 


0.093 


0.088 


0.157 


0.163 


8 


0.573 


0.417 


0.412 


0.708 


0.713 


31 


0.153 


0.106 


0.101 


0.180 


0.185 


16 


0.604 


0.440 


0.435 


0.747 


0.753 


3| 


0.166 


0.116 


0.111 


0.197 


0.202 


1 


0.635 


0.463 


0.458 


0.787 


0.792 


3 


0.177 


0.124 


0.118 


0.210 


0.215 


if 


0.666 


0.487 


0.481 


0.826 


0.831 


2| 


0.198 


0.139 


0.134 


0.236 


0.241 




0.698 


0.510 


0.505 


0.865 


0.870 


21 


0.210 


0.148 


0.143 


0.252 


0.257 


1 fi 

?3 


0.729 


0.533 


0.528 


0.905 


0.910 


2f 


0.229 


0.162 


0.157 


0.275 


0.280 


2 


0.760 


0.556 


0.551 


0.944 


0.949 


2 


0.260 


0.185 


0.180 


0.315 


0.320 




0.823 


0.603 


0.597 


1.023 


1.028 


If 


0.291 


0.209 


0.203 


0.354 


0.359 


% 


0.885 


0.649 


0.644 


1.101 


1.106 














A 


0.948 


0.695 


0.690 


1.18C 


1.185 














1 
2 


1.010 


0.741 


0.736 


1.259 


1.264 



INCLINED PLANE, WEDGE, SCREW, WORM AND WHEEL 185 



SQUARE THREADS 




Fig. 227 

Threads per inch may be about f of the U. S. Standard 
on both Acme and Square. 

189. Single and Multiple Threads. All of the threads shown in 
the preceding illustrations are single threads; that is, the threads are 
formed by the metal left between the successive turns of a single helical 
groove cut around and around the cylinder. If two parallel helical 
grooves are cut, the metal remaining will constitute a double thread; 
three parallel grooves will leave a triple thread, and so on. The single, 
double, and triple threads are illustrated in Figs. 228, 229, and 230, 
respectively. It will be noticed on the single thread (Fig. 228) that 






Single Thread 
Fig. 228 



Double Thread 
Fig. 229 



Triple Thread 
Fig. 230 



if the finger be placed on any point of the thread, as at A, and is moved 
along the thread until it has gone once around the screw, it will come 
to the point C. That is, in moving once around the screw the finger 
has advanced along the screw a distance AC. On the double thread 
(Fig. 229) if the finger starts at A and follows the thread once around, 
it will come to C, but this time there is a point D which lies between 
A and C. D is the point of the second or parallel thread. Similarly, if 
the finger follows a thread in Fig. 230 once around from A to C, two points 
D and E will he between A and C. A multiple thread may be used 
when there is need for a fine thread having a large "lead" (see § 190). 
190. Lead and Pitch of a Screw. The distance AC which the 
thread advances along the screw in one turn around is sometimes 



186 



ELEMENTS OF MECHANISM 



called the pitch. A better name, however, is the lead. This definition 
of lead applies equally to single and multiple threads, while the term 
pitch is usually applied to the distance from one point to the next, 
regardless of the condition of the screw being single or multiple, and 
will be so used in this book. Lead is never used in this sense. In the 
case of a single threaded screw the lead and pitch are the same. If 
a nut is stationary and the screw is turned once around, it will move 
along through the nut a distance equal to the lead. If the screw is 
held from moving endwise but can turn, while the nut is held from 
turning but is free to move along the screw, one turn of the screw will 
move the nut a distance equal to the lead. 

191. Threads per Inch. The size of a thread on a screw is com- 
monly specified by stating the number of threads which the screw has 
in an inch of its length. For example, Fig. 231 represents the side of a 





Fig. 231 



Fig. 232 



screw with a scale laid against it so that the line M is over the center 
of a groove. The fine N, which is an inch from M, comes over the 
center of another groove or another turn of the same groove and there 
are five whole thread points between M and N. This screw would be 
described as a screw having five threads to the inch, no account being 
taken of its being single or double. In Fig. 232 the line M is placed 
over the center of a space and the line N happens to come over the center 
of a thread point with seven whole thread points between. There are, 
therefore, seven and one-half threads per inch on this screw. The num- 
ber of threads per inch is the reciprocal of the pitch and for a single 
threaded screw is also the reciprocal of the lead. 

192. Right-hand and Left-hand Threads. The thread may wind 
around the screw in such a way that it slants downward from right to 
left as one looks at it, as shown in Fig. 233, in which case it is called a 
right-hand thread; or, it may slant downward from left to right as one 
looks at it, as shown in Fig. 234, when it is called a left-hand thread. 
If the screw with the right-hand thread is turned in the direction of 



INCLINED PLANE, WEDGE, SCREW, WORM AND WHEEL 187 



f~^ 


f. — 


A 


/~\ 





















the Arrow A, Fig. 233, it will move downward through the stationary 
nut, or if the screw cannot move endwise the nut will be drawn up. 
The screw with the left-hand thread would have to be turned in the 
direction of the arrow B 
(Fig. 234), to move down- 
ward or to draw the nut 
up. If one were looking 
at the end of a right- 
hand screw and turned it 
right-handed or clockwise, 
it would move away 
from him, whereas a left- 
handed screw looked at 
endwise and turned left- 
handed or anti-clockwise 
would move away. 

193. Relation between the Speed of a Screw or Nut and the Speed 
of a Point on the Wrench or Handle. In Fig. 235 suppose the screw 
S is supported in a bearing. Collars H and B prevent it from moving 
endwise. The lead of the screw is P inches. $ fits into a nut N 
which is free to slide along the guides G which also keep it from turning. 
A crank with a handle K is fast to the end of the screw, the center of K 





Fig. 233 



Fig. 234 






B 






/ 




G 1 


A 




N 
1 




w& 


flf- 


f 








1 



Lead =,P finches 



T 



Fig. 235 



being at a distance of R inches from the axis of the screw. It is now 
required to find a method of determining the relation between the 
linear speed of the handle K and of the nut N. If the crank is given 
one complete turn it will, of course, turn the screw once and the nut 
will move along the guides a distance P inches. While the crank turns 
once the center of K moves over the circumference of a circle whose 
radius is R, therefore it moves over a distance 2 irR inches. Therefore 



Linear speed of N 
Linear speed of K 



P 
2irR 



(62) 



188 



ELEMENTS OF MECHANISM 



Also, since the forces at the two points are inversely as the speeds, 
neglecting friction, 

(63) 



Force at N 



Force at K 



2jR 
P 



In Fig. 236, which shows an ordinary jack-screw, the exact value of 
the speed ratio differs slightly from that expressed by Eq. (62). Here 
the point K at which the force is applied 
rises with the screw so that in making a 
complete turn the point K moves over a 
helix whose diameter is 2 R and whose lead 
is equal to that of the screw. The form ula 
for the length of a helix is v 2 wR 2 




that the actual speed ratio is 



Linear 



of W 



+ P 2 so 



(64) 



Linear speed of K V^rR 2 + P 2 

The lead (P) is so small relative to R 
that the value 



J, 



s, 



V2 ttR 2 + P 2 

differs only very slightly from 2 irR. Ac- 
cordingly, although Eq. (64) is the correct one, Eq. (62) is usually 
accurate enough for all practical purposes. 

194. Compound or Differential Screws. Fig. 237 illustrates the 
style of screw known as a differential screw. A part $ of the screw 
itself has a thread whose lead is P inches and fits into a nut T which 
is a part of the stationary 

frame. The other end S, of ^ f/ ^///Mm^^/mh 

the screw has a different 
thread, of lead P± inches which 
fits the nut N. This nut may 
slide along the guides G but 
is held by the guides from 
turning. As the screw is 
turned the motion of the nut 
is the resultant of the move- 
ment of the screw S through the nut T and of the nut N along Si. Sup- 
pose, for example, that P = \ in. and Pi = re m -> both being right- 
handed screws. If now the handle K is turned right-handed, as seen 
from the left, the whole screw moves along through T toward the right 
\ in. and if it were not for the thread Si N would move to the right \ in. 



Lead^P' 



|<-yv 



B- 



Lead-. P" 



Fig. 237 



INCLINED PLANE, WEDGE, SCREW, WORM AND WHEEL 189 



J 



At the same time, however, Si has drawn N back upon itself tg m - so 
that the net movement of N toward the right is | in. — • ^ in. or ^ in. 
Again, suppose P = | in. right hand and Pi = tg in. left hand. One 
turn of the handle in the same direction as before will advance S through 
T | in. and at the same time carry N off Si ^- in., so that the net move- 
ment of N to the right is \ + yw m - or if m - A device of the first sort 
may be used for obtaining a very small movement of the nut for one 
turn of the screw without the necessity of using a very fine thread. 
195. Examples on Velocity and Power of Screws. 

Example 41. In Fig. 238 suppose it is required to find the load W, which, sus- 
pended from the nut N, can be raised by a force of 60 lb. applied at F. The screw 
has a lead of \ in. Assume that the friction loss is 40 per 
cent. Let R = 20 in. 

Solution. While the screw makes one turn F moves over 
a distance 2 x 20 = 125.66 in. and N rises | in. 
Therefore, F X 125.66 in. = W X h in. 

Since 40 per cent is lost in friction the net force is 
.60 X 60 = 36 lb. 
Therefore, 36 X 125.66 = WX h in., 

or W = 9047.5 lb. 

The same result would be obtained by substituting 
directly in Eq. (63). 

Example 42. In the jack-screw shown in Fig. 236, the 
lead of the screw is § in. R = 3 ft., 6 in. The force 
exerted at K is 100 lb. To find the weight W which could 
be lifted if friction were neglected. 

Solution. Equation (64) applies in this case in finding the speed ratio, 
equation (64) will be very nearly correct. 

Speed of W _ f in. _ 100 
Speed of if ~ 2 x 42 ~ W ' 
Therefore, W = 2 tt 42 X 100 X 2 = 54,779. 

In any case such as this the loss by friction would 
be great and would have to be taken account of. 

Example 43. In Fig. 239 Pi = A in. right hand; 
Pi = | in. right hand. To find how many turns of 
the hand wheel are required to lower the slide \ in., 
and to determine the direction the wheel must be 
turned. 

Solution. Since the outer screw is right hand 
and has a lead of r& in. one turn of the wheel 
right-handed as seen from above will lower the outer 
screw A- in. At the same time, since the inner 
screw is also right-handed, this one turn of the wheel 
will draw the inner screw into the outer one ^ in. so that the resultant downward 
motion of the slide for one turn of the wheel is j^ in. — | in. = t§ in. Therefore, 
to lower it | in. the wheel must be turned right-handed as seen from above as many 
times as y& is contained in | or 8 times. 



^d 


""n 


W^ 


jM 




!/V l| 


J I 




/ - \ 



Fig. 238 



but 




190 ELEMENTS OF MECHANISM 

196. Rotation of Screw or Nut Caused by Axial Pressure. In the 

cases above considered the rotating force has been assumed to act on 
the screw or nut in a plane perpendicular to the axis of the screw. With 
a screw of large lead and relatively small diameter, so that the angle 
which the helix makes with the axis of the screw is small, a force acting 
in the direction of the axis may have a component in the direction to 
cause rotation which is great enough to overcome the frictional resist- 
ance and other resistances to turning and thus cause either the screw 
or the nut to turn. This principle is made use of in small automatic 
drills and screw-drivers, in which axial pressure on the handle causes 
the tool to turn. Such action is not possible unless the helix angle is 
small, and the rotative component of the force relatively large. It is 
well known, however, that constant jarring will cause nuts to work 
loose, hence the necessity for cotter pins or double nuts, one serving 
as a check for the other. 

197. Screw Cutting. Screws are correctly cut in a lathe where the 
cylindrical blank is made to rotate uniformly on its axis, while a tool, 
having the same contour as the space between the threads, is made to 
move uniformly on guides in a path parallel to the axis of the screw 
an amount equal to the lead for each rotation of the blank. The screw 
is completed by successive cuts, the tool being advanced nearer the axis 
for each cut until the proper size is obtained. A nut can be cut in the 
same way by using a tool of the proper shape and moving it away from 
the axis for successive cuts. 

Screws are also cut with solid dies either by hand or power, and with 
proper dies and care good work will result. Nuts are generally threaded 
by means of "taps" which are made of cylindrical pieces of steel having 
a screw-thread cut upon them of the requisite pitch; grooves or flutes 
are made parallel to the axis to furnish cutting edges, the tap is tapered 
off at the end to allow it to enter the nut, and the threads are " backed 
off" to supply the necessary clearance. Before tapping, the nut must, 
have a plain hole in it of a diameter a little greater than the root diam- 
eter of the screw which it is to fit. 

Screws cut by open dies that are gradually closed in as the screw is 
being cut are not accurate, as the screw is begun on the outside of the 
cylinder by the part of the die which must eventually cut the bottom 
of the thread on a considerably smaller cylinder. Thus, as the angle of 
the helix is greater the smaller the cylinder, the lead remaining the same, 
the die at first traces a groove having a lead due to the greater angle 
of the helix at the bottom of the thread. As the die-plates are made to 
approach each other, they tend to bring back this helical groove to the 
standard lead; this strains the material of the threads, and finally pro- 
duces a screw of a different lead than that of the die-plates. 



INCLINED PLANE, WEDGE, SCREW, WORM AND WHEEL 191 



198. Screw Cutting in a Lathe. In cutting a screw thread in a 
lathe, the stock on which the thread is being cut turns at a speed such 
that it will have a surface speed suitable for the cutting tool. While 
the work is making one turn, the tool must be fed along in a direction 




Fig. 240a 

parallel to the axis of the work a distance equal to the lead of the thread 

which is being cut. Figs. 240a and 240b show one of the simplest methods 

of accomplishing this result. 

Fig. 240a is the front view of the 

lathe and Fig. 240b the end view. 

The gears are lettered alike in 

both views. 

Many of the modern lathes use 
a much more elaborate system 
of gearing, but that shown in the 
figure serves to illustrate the 
principles and is easier to under- 
stand than the more complicated 
ones. 

In Fig. 240a, W is the stock on 
which the thread is to be cut. 




Fig. 240b 



This is clamped to the face plate by the dog so that both turn together. 
The face plate is fast to the spindle which is driven from the cone pulley 
either directly or through the back gears. On the opposite end of the 



192 ELEMENTS OF MECHANISM 

spindle is the gear A driving gear B on the stud K through one or two 
idle gears M and N according to the desired direction of rotation. Fast 
to the same stud, and, therefore, turning at the same speed as B, is the 
gear C. This gear drives D through an idle gear. D is fast to the 
lead screw which is embraced by a nut inside the carriage. The tool is 
supported on and moves along with the carriage. 

Assume that the lead of the thread to be cut on the blank is - part 

1 n 
of an inch and that the lead of the thread on the lead screw is - part of 

an inch. If the blank makes a turns in a unit of time, then the distance 

which the tool must move in that time must be a X -; also if b repre- 

n 

sents the number of turns which the lead screw makes in the same unit 

of time, b X - must equal the distance the tool moves. Therefore, 
t 

a X - = b X 7 - 
n t 



Therefore, 



1 

b n 

t 



Angular speed of lead screw _ lead of thread which is being cut . . 
Angular speed of blank lead of thread on lead screw 

Now from the laws governing wheel trains 

Angular speed of lead screw __ teeth in A teeth in C 
Angular speed of blank teeth in B teeth in D 

Therefore, 

Teeth in A teeth in C _ lead of thread which is being cut . . 
Teeth in B teeth in D lead of thread on lead screw 

In any particular lathe the teeth in gears A and B are known quan- 
tities and cannot be changed. 

The lead of the thread on the lead screw is also known. The gears 
C and D can be changed to give the desired speed to the lead screw, 
the idler E being adjusted so as to make proper connection between 
them. If the thread on the lead screw and that being cut are both right 
hand or both left hand, the lead screw must turn in the same direction 
as the blank. If one thread is right hand and the other left hand, the 
lead screw and the blank must turn in opposite directions. This is 
adjusted by the idle gears M and N. 



INCLINED PLANE, WEDGE, SCREW, WORM AND WHEEL 193 

Example 44. In Figs. 240a and 240b, assume that the lead of the thread on the 
lead screw is f in. left hand; gear A has 20 teeth; B, 30 teeth; C, 27 teeth, and D, 
54 teeth. To find the lead of the thread which is being cut on the blank. 

Solution. Substituting in Eq. (66), 

20 27 _ lead of thread being cut 
30 X 54 " | 

Solving this equation gives lead of thread which is being cut as § in. That is, a screw 
of 8 threads per inch is being cut. 

To determine whether a right-hand or a left-hand thread is being cut the direc- 
tions may be followed through by putting on arrows. If this were done in the figure 
the arrows would indicate that the blank and the lead screw are turning in the same 
direction, therefore, since the lead screw has a left-hand thread the thread which is 
being cut is left hand. If the lever R were thrown up so as to bring both idle gears 
into use, the direction of the lead screw would be reversed and a right-hand thread 
would be produced. 

Example 45. Referring still to Figs. 240a and 240b, assume that the lead screw 
and the gears A and B are the same as in Example 44. Let it be required to find 
the number of teeth in C and D to cut 20 threads per inch on the blank. 

Solution. Substituting in Eq. (66), 

20 Teeth in C = fa 
30 Teeth in D f " 
Teeth in C _ 1 8 30 _ 1 
Mence ' Teeth in D ~ 20 X 3 X 20 ~ 5 ' 

Then any practical sized gears may be used at C and D provided D has five times 
as many teeth as C; as, for example, 20 teeth in D and 100 teeth in C. 

199. Worm and Wheel.* Fig. 241 is a picture of a worm and 
wheel mechanism mounted on a frame so as to be used as a model. 
The worm is merely a screw while the wheel is a gear with teeth so 
shaped that they mesh properly into the spaces of the worm thread. 
The worm may be right hand or left hand and single or multiple 
threaded. 

Just as a screw, when turned, moves the nut along, so the worm, when 
turned, pushes the teeth of the worm wheel along, causing the wheel 
to turn. One turn of the worm will move a point on the pitch circle 
of the wheel over an arc equal in length to the pitch or lead of the 
worm. Therefore, in order to cause the wheel to make a complete turn 
the worm must turn as many times as the lead is contained in the 
circumference of the pitch circle of the wheel. 

Let P represent the lead of the worm and D the pitch diameter of 
the wheel. 

r^i Turns of worm tD ,__. 

Then Turns of wheel '?' (67) 

P 

or Turns of wheel = Turns of worm X — ^ • (68) 

* See also Chap. V. 



194 



ELEMENTS OF MECHANISM 



Again, the action of the worm on the wheel may be considered similar 
to the action of a rack on a gear. One turn of a single-threaded worm 
is the same as sliding a rack along a distance equal to the circular pitch 

of the wheel, or turning the wheel through -^ part of a turn, where T 

represents the number of teeth in the wheel. One turn of a double- 



Wonn 




Fig. 241 

threaded worm corresponds to moving a rack along a distance equal 
to twice the circular pitch of the wheel, or turning the wheel through 

j, part of a turn. 

Therefore, 

Turns of wheel = ^ X turns of single-threaded worm, (69) 

2 
Turns of wheel = — X turns of double-threaded worm, (70) 



Turns of wheel = „- X turns of triple-threaded worm. 



(71) 



INCLINED PLANE, WEDGE, SCREW, WORM AND WHEEL 195 



Or, to express the same idea in more general terms, 

Turns of wheet _ Number of threads in worm 
Turns of worm Number of teeth in wheel 



(72) 



It will be noticed that the angular speed ratio of a worm and worm 
wheel as expressed in Eq. (72) is the same as for a pair of gears, consider- 
ing the worm as a gear. In fact the worm and wheel do not differ 
essentially from helical gears, with shafts at right angles, and as has 
been stated in Chapter V it is not easy to determine where the line is 
drawn between helical gears and worm and wheel. Perhaps a general 
method of distinguishing between the two is as follows : 

In helical gears each tooth on each gear is only a part of a helix of 
large lead, whereas in the worm and wheel the wheel teeth are short 
lengths of helices of very great lead while the worm itself is a gear of 
one, two or, at any rate, very few teeth, each of which winds around 
once or more times. 

200. Examples of Worm and Wheel. 

Example 46. In the worm and wheel mechanism shown in Fig. 242 let it be 
required to find the number of turns of the worm that would be necessary to turn 
the wheel 14 times. 

Solution. Applying Eq. (67), 

Turns of worm -w 10 



Therefore 




Circle of 
Wheel 10"dm, 



Example 47. The cylinder C, Fig. 243, is keyed to the same shaft as the worm 
wheel. It is required to find the force F which would be necessary on the handle 
in order to raise the weight of 1000 lb. if friction is neglected. 



196 ELEMENTS OF MECHANISM 

Solution 1. First determine the ratio of the linear speeds of the weight and the 
point at which the force F is applied. If the worm is assumed to make one turn' in 
a unit of time the handle will have a speed equal to the circumference of a circle 
whose radius is the distance from the axis of the worm to the center of the handle. 
Therefore, it would be 2 ir 16 in. = 32 w in. While the worm turns once the wheel 
will turn 

Lead .503 

Circumference of wheel pitch circle tt 12 

Since the cylinder C turns at the same angular speed as the worm wheel a point 

503 
on its circumference will have a linear speed of '— — X it 8 = .3353 nearly. 

IT IZ 

Therefore, since the force is to the weight as the speed of the weight is to the 
speed of the point at which the force is applied, 

F .3353 „ 01 „ 

iwo-w or F = 3 * lb - 

Solution 2. This problem might have been solved by a somewhat shorter method, 
as follows: Assuming the worm single-threaded, the wheel must 'have as many 

teeth as the lead is contained in the circumference of the pitch circle, or -=7-; = 75 

.503 

teeth. Therefore, one turn of the worm will cause the wheel and the cylinder C to 

make J s of a turn (see Eq. 69). Then, if the radius of the crank were equal to the 

radius of C, the force F would be ? V of the weight W. But since the radius of the 

crank is 4 times that of C the force F will be \ X A of W or 3-^. 

Therefore, F= -\%%°- or 3| lbs. 

The same result would be obtained if the worm were not single-threaded, pro- 
vided the lead is the same. For example, assume the worm double-threaded, then 

the wheel would have 2 X vt^' or 150 teeth and one turn of the worm would cause 
.oU3 

the wheel to make T |^ of a turn (see Eq. 70), which equals Js as before. 



CHAPTER IX 



CAMS 

201. A cam is a plate, cylinder, or other solid having a curved out- 
line or a curved groove, which rotates about a fixed axis and, by its 
rotation, imparts motion to a piece in contact with it, known as the 
follower. 

This motion may be transmitted by sliding contact ; but where there 
is much force transmitted, it is often by accomplished rolling contact. 

If the action of the piece is intermittent, it is sometimes called a 
wiper ; that is, a cam, in most places, is continuous in its action, while 
a wiper is always intermittent : 




Ui 



L 



but a wiper is often called a 
cam notwithstanding. 

Fig. 244 is a drawing of a 
cam known as a plate cam, and 
Fig. 245 a drawing of a cylinder 
containing an irregular groove 
and known as a cylindrical 
cam. 

Very many machines, parti- 
cularly automatic machines, 
depend largely upon cams, 
properly designed and properly 
timed, to give motion to the 
various parts. 

Usually a cam is designed 
for the special purpose for 
which it is to be used. In most 
cases which occur in practice 
the condition to be fulfilled 

in designing a cam does not directly involve the speed ratio, but assigns 
a certain series of definite positions which the follower is to assume 
while the driver occupies a corresponding series of definite positions. 

The relations between the successive positions of the driver and fol- 
lower in a cam motion may be represented by means of a diagram 
whose abscissae are linear distances arbitrarily chosen to represent angu- 

. 197 



Fig. 244 



198 



ELEMENTS OF MECHANISM 



lar motion of the cam and whose ordinates are the corresponding dis- 
placements of the follower from its initial position. This is illustrated 
in Fig. 246, where the line Oabc represents the motion given by the cam. 




Fig. 245 

The perpendicular distance of any point in the line from the axis OY 
represents the angular motion of the driver, while the perpendicular 
distance of the point from OX represents the corresponding movement 
of the follower, from some point considered as a starting point. Thus 
the line of motion Oabc indicates that from the position to 4 of the 
driver, the follower had no motion; from the position 4 to 12 of the 
driver, the follower had a uniform upward motion 612; and from posi- 




Fig. 246 

tion 12 to 16 of the driver, the follower had a uniform downward motion 
612, thus bringing it again to its starting-point. 

201A. Diagrams for Cams giving Rapid Movements. It is very 
often the case that a cam is required to give a definite motion in a 
short interval of time, the nature of the motion not being fixed. The 
form of the diagram for such a motion will now be discussed. 

In the diagram shown in connection with Fig. 246 the follower had 
two uniform motions, and if the cam be made to revolve quickly, quite 
a shock will occur at each of the points where the motion changes, as 
a, 6, and c; to obviate this the form of the diagram can be changed, 
provided it is allowable to change the nature of the motion. 



CAMS 



199 




Fig. 246a 



Suppose a cam is to raise a body rapidly from e to/ (Fig. 246a), the 
nature of the motion to be such that the shock shall be as light as 
possible. 

For the straight line Oa the case is one of a uniform motion (as in Fig. 
246), the body being raised from e to / in an interval proportional to 
ob; here the motion changes suddenly at and a accompanied by a 
perceptible shock. The line Ocda would be an improvement, the fol- 
lower not requiring so great 
an impulse at the start or 
near the end of the motion 
each being much more 
gradual than before. 

The body may be made to 
move with a harmonic 
motion, the diagram for which 
would be drawn as follows 
(Fig. 246b): 

Draw the semicircle ebf on ef as a diameter; divide the time line 
Oh into a convenient number of equal parts (in this case ten), and then 
divide the semicircle into the same number of equal parts; through the 
divisions of the semicircle draw horizontal lines intersecting the vertical 
lines drawn through the corresponding points of division of the time 
line Oh, thus obtain- 
ing points, as a, b, c, 
etc. A smooth curve 
drawn through 
these points gives 
the full curve Oabcd 
. . . n. Here the 
body or follower 
receives a velocity 
increasing from zero 
at the start to a 
maximum at the middle of its path, when it is again gradually dimin- 
ished to zero at /, the end of its path. 

This form of diagram gives very good results, and is satisfactory in 
many of its practical applications. 

A body dropped from the hand has no initial velocity at the start, but 
has a uniformly increasing velocity, under the action of gravity, until it 
reaches the ground; similarly, if the body is thrown upward with the 
velocity it had on striking the ground, it will come to rest at a height 
equal to that from which it was dropped, and its upward motion is the 




Fig. 246b 



200 



ELEMENTS OF MECHANISM 



reverse of the downward one, that is, a uniformly retarded motion. 
(See § 27.) 

In designing a cam for rapid movement the motion of the follower 
should obey the same law of gravity, and have a uniformly accelerated 
motion until the middle of its path is reached, then a uniformly retarded 
motion to the end of its path. 

A body free to fall descends through spaces, during successive units 
of time, proportional to the odd numbers 1, 3, 5, 7, 9, etc., and the 
total space passed over equals the sum of these spaces. 

To develop a line of action according to this law upon the same time 
line Oh, and with the same motion ef, as before, proceed as follows: 

Divide the time line Oh into any even number of equal parts, as ten; 
then divide the line of motion ef into successive spaces proportional to 
the numbers 1, 3, 5, 7, 9, 9, 7, 5, 3, 1, and draw horizontal lines through 
the ends of these spaces, obtaining the intersections a', b', c', etc., with 
the vertical lines through the corresponding time divisions 1, 2, 3, etc.; 
a smooth curve, shown dotted in the figure, drawn through these points, 
will give the cam diagram. 

202. Plate Cams. A plate cam imparts motion to a follower guided 
so that it is constrained to move in a plane which is perpendicular to 
the axis about which the cam rotates; that is, in a 
plane coincident with or parallel to the plane in which 
the cam itself lies. The character of the motion given 
to the follower depends upon the shape of the cam. 
The follower may move continuously or inter- 
mittently; it may move with uniform speed or 
variable speed; or it may have uniform speed part 
of the time and variable speed part of the time. A 
knowledge of the various types of plate cams, and an 
idea of the manner of attacking the problem of 
designing a cam for any specific purpose, can best be 
obtained by studying a number of examples. 




Cam Shaft 



Example 48. A cam is to be keyed to the cam shaft (Fig. 
247), which turns as indicated. The shape of the cam is to be 
such that the point of the slider S will be raised with uniform 
motion from A to B while the cam makes one-half a turn, and 
lowered again to the original position during the second half- 
turn of the cam. The cam shaft turns at uniform speed. 
Solution. (Fig. 248.) Draw a circle through A with C as a center. Since the 
follower is to rise from A to B while the cam makes one-half a turn (or turns through 
180°), and since the cam shaft turns at uniform speed, divide one-half of the circle 
(AVW) into any number of equal angles by the lines Ca, Cb, Cd, and Ce. Four 
divisions are made in the illustration, although for accurate work a greater number 



CAMS 



201 



would be desirable. The divisions are made on the side which is turning upward 
toward the follower, that is, back on the side/rom'which the arrow is pointing. Now, 
divide the distance AB into as many parts as there are divisions in the angle A7W. 
Since the follower is to rise from A to B with uniform motion, the divisions of AB 
will be equal. That is, A to 1 = 1 to 2 = 2 to 3 = 3 to B. When the cam has 




Fig. 248 



made one-fourth of a half revolution, the line Ca will be vertical. A point m on 
this line, found by swinging an arc through 1 with center C, will be the point on the 
cam which will be at the height CI above the center when the cam has made one- 
fourth of one-half revolution. Similarly, n will be the point on the cam which will 
be at 2 when the cam has turned one-half of the half revolution, p and r are found 
in the same way, by drawing arcs through 3 and B cutting the lines Cd and Ce, 
respectively. 



202 



ELEMENTS OF MECHANISM 



A smooth curve drawn through the points A, m, n, p, and r will be the correct 
outline for that portion of the cam which will raise the follower point from A to B 
as specified. Since the follower is to be lowered from B to A, also, with uniform 
motion during the remaining half turn of the cam, the other hah of the cam outline 
will be a duplicate of that already found. 

Example 49. Data the same as for Example 48, except that the follower, instead 
of having a point shaped as in that case, has a roller, as shown in Fig. 249, on which 
the cam acts. The construction 
is shown in Fig. 250. It is neces- 
sary first to find the outline of the 
cam for a follower like that in Fig. 
248, the point of the follower 
being assumed to be at the center 
A of the roller, Fig. 250. The 
construction of this curve is 
exactly the same as explained for 
Fig. 248 and is lettered the same 
in Fig. 250, the curve itself being 
drawn as a dot and dash line. 
This is called the pitch line of the 
cam. The next step is to set a 
compass to a radius equal to the 
radius of the roller and, with 





Fig. 249 



Fig. 250 



centers at frequent intervals on the pitch line, draw arcs as shown dotted. The 
true cam outline is a smooth curve drawn tangent to these arcs. It should be noted 
that the point of tangency will not necessarily lie on the line joining the center of 
the arc to the center of the cam. For instance, consider the arc drawn with n as 
a center. The cam curve happens to strike this arc at y, not at the point where the 
arc cuts the line Cb. 



CAMS 



203 



This condition often prevents the cam which acts on a roller or similar follower 
from giving exactly the same motion as would be obtained from the "pitch line" 
cam acting on a pointed follower. This is likely to be true 
at convex places where the motion changes suddenly. 

Example 50. Given a follower with a roller as shown 
in Fig. 251. The lowest position of the center of the 
roller is a distance N above the center of the cam shaft, 
and the line AB along which the center of the roller is 
guided is a distance D to the right of a vertical line 
through C. That is, the center of the cam shaft is 
offset a distance D to the left of the line of motion of 
the center of the follower. To draw the outline of a 
plate cam which, by turning as shown by the arrow, 
shall raise the center of the roller from A to B with 
uniform motion while the cam makes one half a turn, 
then lower it again to A during the second half revolu- 
tion of the cam. 

Solution. Fig. 252 shows the solution of this problem. 
Starting with C, locate the center A by measuring a 
distance D to the right of C and a distance N above C. 
Draw a line Ck through C and A. Since the upward 
motion is to take place during one-half turn of the cam, measure back 180° from Ck 
and draw Ce (that is, kACe is a straight line). Divide the angle kCe into any 




am Shaft 



Fig. 251 




convenient number of equal parts as before (in this case four) by the lines Ca, Cb, 
Cd. Divide AB into the same number of equal parts, since the follower is to rise 



204 



ELEMENTS OF MECHANISM 



with uniform speed. From C as a center swing an arc through 1 cutting Ck at 5. 
Cut Ca with the same arc at 9. Make the length 9-10 equal to 5-1. Then 10 is one 
point on the pitch line of the cam. In the same way point 12 is found by making arc 
11-12 equal to arc 6-2, and, similarly, all the way around. The true cam outline is 
found as before by drawing arcs with radius equal to the radii of the roller, and with 
centers on the pitch line, and then drawing a smooth curve tangent to these arcs. 

Example 51. Fig. 253 shows the method of laying out a cam to move a follower 
from A to B with uniform motion during one-quarter turn of the cam, hold it at B 




Fig. 253 



during one-quarter turn, lower it again to A during one-quarter turn, and allow it 
to remain at A during the last quarter turn; the cam to turn in the direction of the 
arrow. 

Solution. Starting with C and A as in the preceding figure, draw the line kAC. 
It is convenient, for the purpose of dividing up the angles, to draw a circle through 
A with C as a center. With CAk as a starting or reference line divide the circle into 
four equal parts by the lines Ce, Cf and Cn. The angle kCe is the angle through 
which the cam turns while the follower is being raised. Divide this angle into equal 
parts, and find points on the cam pitch line as described in Example 50, the only 
difference here being that the angles kCa, kCb, etc., are smaller. Point 16 is the last 
point thus found. Since the follower is to remain at rest at B during the next quarter 
turn of the cam, the pitch fine of the part of the cam which holds it up there will 
be an arc of a circle drawn through 16 with C as a center. This part of the cam will 



CAMS 



205 



extend from 16 to, 18, the point 18 being found by extending the circle to cut Cf at 
17 and making the distance 17-18 equal to 15-16. The construction for completing 
the pitch line is exactly similar, and the cam curve proper is obtained as described 
in the previous examples. 

Example 52. Fig. 254 is a cam which raises the center of the roller from A to 
B with harmonic motion during one-third of a turn, allows it to drop to its original 
position instantly, and holds it there during the remaining two-thirds of a turn. 
The angle kCf, through which the cam turns to raise the roller, is laid off (120°) and 
divided into an even number of equal parts. Since the roller is to rise with harmonic 
motion, a semicircle is drawn with AB as a diameter, and the circumference of this 
semicircle is divided into as many equal parts as there are divisions in the angle kCf. 




Fig. 254 



Fig. 255 



From the points of division on this semicircle perpendiculars are drawn to the line 
AB, meeting it at points 1, 2, 3. These points are the points of division of AB to 
be used in finding the pitch line of the cam, which is found as previously described. 
The last point on the part which raises the follower is 16. Since the follower is to 
drop instantly, draw a straight line from 16 to 17, the point where an arc through 
A cuts Cf. The remainder of the pitch line is a circle about C through 17 around 
to A. 

Example 53. In Fig. 255 a cam is to be placed on the shaft at C to act on a roller 
centered at A on the rocker ART. Another roller centered at T on the rocker fits 
a slot in the slider S. The cam is to be of such shape that, by turning as shown 
by the arrow, it will move the slider with harmonic motion to the position shown 



206 



ELEMENTS OF MECHANISM 



1 - 




<< \ 



Fig. 256 



CAMS 



207 



dotted during one-half turn of the cam in the direction indicated, allow it to return 
to its original position, with harmonic motion, during the next 1 of a turn and allow 
it to remain at rest during the remaining | of a turn. Fig. 256 shows the con- 
struction. 

Example 54. In Fig. 257 let it be required to design a cam to be placed on shaft 
to raise slider A to Ai during i of a turn of the cam, allow it to drop at once to its 
original position and remain there during the rest of the turn of cam, the charac- 
ter of the motion of A to be unimportant except that the starting and stopping 





Fig. 257 



shall be gradual. The cam is to act on a roller on the rocker BCD, the rocker being 
connected to the slider by the link BA. 

Solution. Fig. 258. First draw the motion diagram assuming uniform motion 
for A. This is shown by the dotted line ta s . Next substitute for this line the line 
shown full, the greater part of which is straight, having more slope than the original 
line and connected to points t and a s by curves drawn tangent to the sloping line 
and tangent to horizontal lines at t and a&. 

Subdivide into equal parts the distance tt 8 , which represents the i turn during 
which the motion of the slider takes place, erect ordinates at these points cutting 
the motion plot at points ai, a% etc., and project these points on to the path of A 
getting Ai, A 2 , etc. From A\, A 2 , etc.,;with radius AB cut an arc drawn about C 
with radius CB, getting B h B 2 , etc. From these points draw lines through C cutting 
the arc of radius CD at D h D 2 , etc. The cam is found from these points as in previous 
examples. 



208 



ELEMENTS OF MECHANISM 




Fig. 258 



CAMS 



209 



203. Positive Motion Plate Cams. It will be noticed that in each 
of the cams which have been discussed, the follower must be held in 
contact with the surface of the cam by some external force such as 
gravity, or a spring. The cam can only force the follower away from 
the cam shaft, while some outside force must bring it back. In case 
it is desired to make the cam positive in its action in either direction 
without depending upon external force, the cam must be so constructed 
as to act on both sides of the follower's roller, or there must be two 
rollers, one on either side of the cam. Fig. 259 shows a cam designed 




Fig. 259 



to give the same motion to the same follower as in Fig. 256. In Fig. 
259, however, the pitch line of the cam is made the center line of a 
groove of a width equal to the roller diameter, thus enabling the cam to 
move the roller in either direction. 

Fig. 260 shows another style of positive motion cam. The follower 
consists of a framework carrying two rollers, one, roller C, resting on 
cam A, which is designed so as to give whatever motion is desired for 
the follower. The other, roller D, rests on cam B, which is designed to 
be in contact with roller D, the position of the latter depending in turn 
upon the position of the roller C. It would be possible to have both 
rollers touching the same cam, but in that case the movement of the 



210 



ELEMENTS OF MECHANISM 



follower could only be chosen for one-half a turn of the cam, the other 
half being determined by the shape of the cam necessary to be in con- 
tact with both rollers. 




Fig. 260 



204. Plate Cam with Flat^Follower. — Example 55. The follower 
for the cam shown in Fig. 261 has a flat plate at its end instead of a 
roller. The cam is so designed that, when it turns right-handed, the 
follower is raised with harmonic motion while the cam makes one-third 
of a turn, then remains at rest during the next third of a turn of the 
cam and is lowered with harmonic motion during the remaining third 
of a turn. 



CAMS 



211 



If the center line of the guides in which the follower moves does not 
pass through the center of the cam, the shape of the cam is not affected, 
provided the direction is the same. 

Solution. The method of construction is as follows : Assuming that the follower 
is shown in its lowest position, measure up along a vertical line passing through the 
center of the cam the distance 08 which the follower is to move. Divide this into 
any even number of harmonic divisions, eight being used in the drawing. Lay back 
the angle oEm equal to the angle through which the cam turns while the follower 
is being lifted. Divide oEm into as many equal angles as there are harmonic divi- 
sions in the line 08. Through point 1 swing an arc with E as a center cutting the 
first radial line at w; through w draw a line perpendicular to Ew. Through 2 draw 




Fig. 261 



an arc cutting the second radial fine at v and draw through v a line perpendicular 
to Ev. In a similar way draw perpendiculars to Eu, Et, Er, Ep, En, and Em. A 
smooth curve tangent to all of these perpendiculars will be the outline of that por- 
tion of the cam which raises the follower. 

Since the follower is to remain at rest while the cam turns through the next 120° 
the outline between the line Em and the line Ek 120° away from Em will be an arc 
of a circle through m with E as a center. 

The outline of the portion of the cam which lowers the follower is found in a 
manner similar to that described for raising it. 

If the foot of the follower made an angle 6 with the center line of its path, 6 being 
other than 90°, the construction lines at c, d, e, f, etc., instead of being drawn per- 
pendicular to Ec, Ed, Ee, Ef, etc., would be drawn making an angle with these lines. 



212 



ELEMENTS OF MECHANISM 



205. Plate Cam with Flat Rocker. — Example 56. The cam in 

Fig. 262 actuates the follower S through the rocker R which is pivoted 
at P. S slides in guides, and remains still while the cam makes a quar- 
ter turn right-handed, then rises to the upper dotted position with 
harmonic motion during a quarter turn of the cam. During the next 




Fig. 262 

quarter turn the follower drops with harmonic motion to its original 
position, and remains at rest during the last quarter turn. The foot of 
the follower is a semicircle with center at o, resting on the upper flat 
surface of the rocker. 

To find the cam outline, first divide the distance oQ into harmonic spaces, six 
being used in this case. These points of divisions are the successive positions of the 
center of the semicircle. Draw arcs of the circle with each of the points, 1, 2, 3, 4, 
5, 6, as centers. Draw the dotted circle K tangent to the upper surface of the 
rocker produced. Next, draw the lines a, b, c, d, e, and / tangent to circle K and 
to the arcs drawn at 1, 2, 3, 4, 5, and 6 respectively. Parallel to, and at a distance 
T from lines a, b, c, etc., draw lines g, h, i, j, etc., cutting the vertical line through 
the cam center C at 7, 8, 9, 10, 11, and 12. 



CAMS 213 

Since the follower is to remain at rest during a quarter turn of the cam, the out- 
line of the cam over the angle A is an arc of a circle with radius CE. 

Since the upward movement takes place during a quarter turn, or 90°, lay off 
angle B equal to 90° and divide it into as many equal angles as there are harmonic 
divisions in oQ. Lay off C14 equal to C7 and through point 14 draw a line making 
the same angle with C14 that line g makes with CE. Draw similar lines through 
each of the other radial lines C15, C16, CT7, CIS, and C19. The cam outline will 
be a smooth curve tangent to all the lines which have been thus drawn. 

A similar construction is used for finding the curve for the part of the cam which 
lowers the follower. The last part of the cam, over angle F, will be a circular arc 
to give the period of rest. 

206. Cylindrical Cams. — Example 57. The general appearance of 
a cylindrical cam has already been shown (see Fig. 245.) Fig. 263 gives 
dimensions for the hub and groove for a 
cylindrical cam which is to hold a follower 
still for one-eighth turn of the cam, move it 
1 in. to the right in a line parallel to the axis 
of the cam, with uniformly accelerated and 
uniformly retarded motion (see §201A) 
while the cam makes three-eighths turn, 
hold it still for one-eighth turn, and return 
it to its original position with similar motion 
in three-eighths turn. 

Solution. The solution of this problem is shown in Fig. 264. The 
upper left-hand view is an end view of the cam, the upper right-hand 
view is a side elevation of the cam. 

To make the drawing, proceed as follows: 

Locate the center line XX'. On the line XX' choose the point C at any con- 
venient place and draw the circle K whose radius is equal to the outside radius of 
the cylinder. Also draw the dotted circle P with the radius equal to the outside 
radius minus the depth of the groove. Draw the vertical center line YY'. Lay 
back the angle YCB equal to | of 360°, that is, 45°. This is the angle through 
which the cam will turn before the follower starts to move. Since the movement 
of the follower is to take place during the next three-eighths of a turn, the cam will 
turn through the angle BCY' to give the motion to the follower. Since the follower 
is to remain at rest during the next one-eighth turn, the angle Y'CT equal to 45° 
will next be drawn, and the remaining angle TCY will be the angle through which 
the cam will turn to move the follower back to its original position. Now, draw 
the center line MN at any convenient distance on the right of the figure already 
drawn, and locate the point E on this line at a distance from XX' equal to the out- 
side radius of the cylinder. On a horizontal line drawn through E locate the points 
F and G, each at a distance from E equal to the radius of the roller on which the 
cam is to act. Draw HJ parallel to FG at a distance from it equal to the depth of 
the groove. Through F and G draw lines to the point L where MN intersects the 
axis XX'. That portion of the line HJ intersected between FL and GL will be the 




214 



ELEMENTS OF MECHANISM 



width of the groove at the bottom. Before it is possible to proceed further in the 
construction of this side elevation of the cam, it is necessary to make a development 
of its outer surface. Draw the line M'N' equal in length to the circumference of the 
cylinder. 




Lay off M'B' equal to the length, of the arc YB and B'Y'i equal to the length of 
the arc BY'. Divide B'Y'% into any even number of equal parts, in this case eight, 
and letter points of division a', b', c', d', e', f, and g'. Through the points thus 
found draw vertical lines. On the vertical line through M' lay off M'8 equal to 



CAMS 215 

the distance through which the follower is to move, and divide M'8 into "gravity" 
divisions (see § 201A), using as many divisions as there are equal divisions in B'Y\. 
Mark the points thus found 1, 2, 3, 4, 5, 6, 7. From 1 project across to the vertical 
through a'. From 2 project to the vertical through b', and so on, thus getting the 
points 9, 10, 11, 12, 13, 14, 15, and 16. A smooth curve drawn through these points 
will be the development of the center line of that portion of the cam groove which 
moves the follower to the right. Make Y' 2 T' equal to the length of the arc Y'T. 
The development of the center line of the groove between the verticals at Y\ and 
T' is a horizontal straight line. Since the return motion of the follower is a dupli- 
cate of the forward motion, the curve 17N', being a duplicate of the curve B'16, 
will be the development of the center line of that portion of the cam groove which 
moves the follower back to its original position. 

The above construction gives a development of the center line of the groove on 
the outer surface of the cylinder. The lines forming the development of the sides 
of the groove are smooth curves drawn tangent to arcs, swung about a series of 
centers along the line M'B' 16 17 N' with radii equal to the radius of the large end 
of the roller as shown in the drawing. Similar curves drawn tangent to arcs swung 
about the same centers with a radius equal to the radius of the large end of the 
roller plus the thickness of the flange forming the sides of the groove, will be the 
development of the outer edges of these flanges. 

The development of the corners of the bottom of the groove is constructed in the 
same way, except that the length of the development is less, because it is a develop- 
ment of a cylinder of smaller radius. 

The projections (on the side elevation) of the curves which have just been de- 
veloped are drawn by finding the projections corresponding to points r', s', t', v', 
where these curves cut the vertical line, it being borne in mind that the vertical lines 
on the development really represent the developed positions of elements of the cylin- 
der, drawn through points a, b, c, etc., which are found by dividing the arcs BY' 
and TY into divisions equal to the divisions in B'Y'z and T'N'. The construction 
for the points r', s', t', and v' only will be followed through as the construction for all 
other points will be exactly similar. Through b on the end view draw an element 
of the cylinder across the side elevation. From e, where this element intersects 
MN, lay off et equal to b't', ev equal to b'v', to the right of MN since t' and v' are above 
M'N', and es equal to b's' and er equal to b'r', to the left since s' and r' are below 
M'N'. The points r, s, t, v, are the projections of points corresponding to r', s', t', v'. 
Projections of all other points where the curves intersect the verticals on the develop- 
ment are found in exactly the same way, and smooth curves drawn through the 
points thus found will be the projections of the corners of the groove, and of the 
flange enclosing the groove. The projections of the corners of the bottom of the 
groove are obtained in the same way also, using, of course, elements through a 2 , b 2 , 
etc., instead of a and 6. 

207. Multiple-turn Cylindrical Cam. Fig. 265 shows a cylindrical 
cam which requires two revolutions to complete the full cycle of motion 
of its follower. The method of designing such a cam would be similar 
in principle to that described for the simple cam in Fig. 264. The fol- 
lower in a case like this may require a special form in order to pass 
properly the places where the groove crosses on itself. This is sug- 
gested in Fig. 266. The follower F is made to fit the groove sidewise, 



216 



ELEMENTS OF MECHANISM 




1 


\\ 

\ I 

\\\ / ' 
\\\ y / 





\\ [Tl 

\ \ 

\\ 

\ ^X/s 

f\ 



CAMS 



217 



and is arranged to turn in the sliding rod, to which it gives motion in a 
line parallel with the axis of the cam. The guides for this rod are 
attached to the bearings of the cam, A and B, which form a part of the 
frame of the machine. A plan of the follower is shown at G: its elon- 
gated shape is necessary so that it may properly cross the junctures of 




the groove. In this cam there is a period of rest during one-half a turn of 
the cam at each end of the motion; the motion from one limit to the other 
is uniform, and consumes one and one-half uniform turns of the cam. 

The cylinder may be increased in length, and the groove may be made 
of any desirable lead; the period of rest can be reduced to zero, or in- 
creased to nearly one turn of the cam. A cylindrical cam, having a 
right- and a left-handed groove, is often used to produce a uniform re- 
ciprocating motion, the right- and left-handed threads or grooves passing 
into each other at the ends of the motion, so that there is no period of rest. 






Fig. 267 



The period of rest in a cylindrical cam, like that shown in Fig. 266, 
can be prolonged through nearly two turns of the cylinder by means 
of the device shown in Fig. 267. A switch is placed at the junction of 



218 ELEMENTS OF MECHANISM 

the right- and left-handed grooves with the circular groove, and it is 
provided that the switch shall be capable of turning a little in either 
direction upon its supporting pin, while the pin is capable of a slight 
longitudinal movement parallel with the axis of the cylinder. This 
supporting pin is constantly urged to the right by a spring, shown in A, 
which acts on a slide carrying the pin; when in this position the space 
a between the switch and the circular part of the groove is too small to 
allow the follower to pass, and when the follower is in the position 
shown in B, the spring is compressed; then, if the follower moves on, 
the space behind it is closed, as the spring will tend to push the sup- 
port to the right, and swing the switch on the follower as a fulcrum. 

If the cam turns in the direction of the arrow, in A the shuttle- 
shaped follower is entering the circular portion of the groove, and leaves 
the switch in a position which will guide the follower into the circular 
groove when it again reaches the switch; in B the switch is pressed 
toward the left to allow the follower to pass. As motion continues, the 
support of the switch is pressed to the right, and the switch is thrown 
into the position shown in C ready to guide the shuttle into the return- 
ing groove. The period of rest, in this case continues for about one 
and two-thirds turns of the cylinder. 

Fig. 268 shows an arrangement which may be applied for guiding a 
wire or cord as it winds upon a spool. The hub of the sheave is bored 
to fit the outside of the shaft. The shaft is stationary and has a right- 




Fig. 268 

handed groove and a left-handed groove cut in it, and is therefore a 
stationary cylindrical cam. On the side of the sheave is a projection 
which supports the pin on which the specially constructed follower is 
carried. The wire or cord, passing over the sheave, causes it to turn, 
and as it turns it receives a reciprocating motion along the axis of the 
cam. 



CAMS 



219 



208. Cylindrical Cam Acting on a Lever. If the follower for a cylin- 
drical cam is a pin or roller on the end of a lever, so that it moves in 
an arc instead of a straight line, as in Fig. 269, an exact construction 
would require that allowance be made for the curvature of the path 
when making the development. This degree of refinement is usually 
unnecessary, from a practical point of view, and the cam may be de- 
signed on the assumption that the path of the follower is a straight line 
parallel to the elements of the cylinder. 

If the lever is in a plane passing through the axis of the cam, as in 
Fig. 270, the end of the lever may be considered as one tooth of a worm 




Fig. 269 



Fig. 270 



wheel or helical gear, and be given the form of such a tooth. The cam 
itself then corresponds to the worm or to the mating helical gear except 
that its groove is not necessarily helical. 

209. Combinations of Two or More Cams. In various automatic 
machines the movements of parts which have to be timed with respect 
to each other'are often obtained by the use of two or more cams properly 
designed, and properly adjusted to give each piece its desired motion 
at the required time. Fig. 271 shows how a cylindrical cam and a 
plate cam might be arranged to work in combination with each other. 
In this case the cylindrical cam makes two revolutions for every one of 
the plate cam. The cylinder R is caused to swing back and forth by 
the lever $ which, in turn, is operated by the plate cam. 

With the mechanism in the position shown, the cylindrical cam makes 
one-eighth turn in the direction shown, after which the pin T starts to 
move to the right with harmonic motion. T moves to the right the total 
distance of If in., during three-eighths of a turn of the cylindrical cam, 
after which it remains at rest for one-eighth turn of the cam, then re- 



220 



ELEMENTS OF MECHANISM 



turns to its original position during the remaining three-eighths turn. 
The plate cam is so designed that, turning left-handed as shown, the 
cylinder R begins to turn after T has moved to the right f of an inch. 




Fig. 271 



It continues to turn with uniformly accelerated and uniformly retarded 
motion until T gets back again to within f of an inch of its left-hand 
position. 

The hole W will then be in the position now occupied by the hole V. 
R will then stop its motion and T will be inserted into the hole W. 
During the next revolution of the cylindrical cam T has a motion the 
same as before, and the plate cam swings the cylinder R back to its 
original position. 



CHAPTER X 

FOUR-BAR LINKAGE. RELATIVE VELOCITIES OF RIGIDLY 
CONNECTED POINTS 

210. The Four-Bar Linkage will be discussed fully in a later chap- 
ter, the purpose of the present chapter being to study the relative linear 
velocities of connected points, particularly points on a linkage. Only 
such consideration will be given to the linkage itself at this time as to 
make it possible to study the velocities understanding^. 

Fig. 272 shows in a simple form a mechanism known as a four-bar 
linkage. E is a fixed piece, such as the frame of a machine. A 




Fig. 272 

and D are shafts having their bearings in E. The line joining the 
centers of A and D is called the line of centers. The piece F, called 
a crank, is keyed to A . H is a similar crank keyed to D. The outer 
ends of F and H are connected to each other by the connecting rod K 
and the crank pins B and C. B may be made fast to K and be free 
to turn in the hole in F or it may be fast to F and free to turn in K. 
Similarly, the pin C may be free to turn in either H or K. 

If shaft A is caused to revolve, the crank F will revolve with it, the 
center of the pin B moving in a circle whose center is the center of A. 
This movement of B will, through the connecting rod K, cause the 

221 



222 



ELEMENTS OF MECHANISM 



pin C to move, and since C can move only in a circle about the center 
of D the crank H will be caused to turn, turning D with it.* Each 
B J< C 



£ 





Fig. 273 

one of the pieces, E, F, K, and H, is called a link, and the whole 
system is called a four-bar linkage. 

It is convenient, in studying linkages, to indicate them by the center 
lines of the links, as shown in Fig. 273 which represents the same linkage 
as that shown in Fig. 272. 

211. Four-Bar Linkage with a Sliding Member. In Fig. 274, the 
end of the connecting rod carries a block, pivoted to it at C, which slides 



1 




mm ia 

11 m ^ 


li 


k 


um i_{iii_jmw 




Fig. 274 

back and forth in the circular slot as the crank AB revolves. The 
center of the slot is at D. The center of the crank pin C evidently has 
the same motion that it would have were it guided by a crank of length 
* Some provision is necessary for passing the positions where the connecting rod 
and driven crank come into line with each other. 



FOUR-BAR LINKAGE 



223 



DC turning about D. The mechanism, therefore, is really a four-bar 
linkage with the lines AB and DC as center lines of the cranks, AD as 
the line of centers, and BC as the center line of the connecting rod. 

Let it now be supposed that the slot is made of greater radius than 
that shown in the figure, for example, with its center at D\. Then the 
equivalent four-bar linkage would be ABCDi. 

Carrying the same idea still further, let the slot be made straight. 
Then the equivalent center D would be at a point an infinite distance 




Fig. 275 

away. The mechanism, however, would still be the equivalent of a 
four-bar linkage, as shown in Fig. 275, where AB is one crank, the line 
through C perpendicular to the slot is the other crank, BC the con- 
necting rod, and a line through A parallel to the crank through C is 
the line of centers. 

Fig. 276 shows the special form in which this linkage commonly 
occurs, where the center line of the slot passes through the center of 




Fig. 276 

the shaft A. This is the mechanism formed by the crank shaft, crank, 
connecting rod, crosshead and crosshead guides of the reciprocating 
steam engine. 



224 



ELEMENTS OF MECHANISM 



212. Relative Motions of the Links. In the four-bar linkage shown 
in Fig. 277, A and D are the stationary axes, AB and DC the cranks 
and BC the connecting rod. If the crank AB turns from the position 
shown in full lines to the right-hand dotted position — that is, turns 
through the angle BABi — the pin B will travel over the arc BBi. This 
will cause the connecting rod to move and push the pin C along its 
path to C\. 

It is apparent from the figure that the length of the arc CCi is not 
equal to the length of the arc BBi. In other words, with the several 
links having the relative lengths as here shown, the linear speeds of 




Fig. 277 



the crank pins will differ. If, furthermore, B is moved to B 2 , C will 
move to C 2 . Now the arc BB 2 is made equal to BB h but CC 2 is evi- 
dently not equal to CCi. This construction, therefore, suggests that 
if the crank A B is turned with uniform angular speed so that the crank 
pin B has a uniform linear speed, the crank pin C has a varying linear 
speed and the crank DC a varying angular speed. It will be shown, 
later (see § 227), that the motions of the links of a specific four-bar 
linkage relative to each other are always the same whichever of the four 
links is the stationary one. 

It will be apparent also, when the preceding statement is shown to 
be true, that the relative linear velocities of points on any of the links are 
independent of the fixedness of the links. 

213. Graphical Representation of Motions and Linear Velocities. 
For convenience in graphical work in connection with the study of 
velocities it is customary to represent the velocity of a point by a 
straight line whose direction and length indicate the direction and 
magnitude of the velocity of the point. For example, let it be assumed 
that the block M (Fig. 278) is sliding to the right on the guide, at the 
rate of one foot per second. If it is desired to represent the velocity of 
any point A on this block by a line, any unit of length may be assumed 



FOUR-BAR LINKAGE 



225 



Guide 
Fig. 278 



to represent one foot per second. Suppose one inch equals one foot per 
second is chosen as the convenient scale; then a line one inch long is 
drawn from A to the right, parallel to 
the guide, with an arrow head at its end 
pointing to the right. 

If the point is moving in a curved 
path the line representing its velocity 
is drawn tangent to the path at the 
position of the point on the instant under consideration. 

If the point has a variable speed the length of the line representing 
its velocity is made equal to the distance the point would move if it 
continued for a unit of time with the same speed which it has at the 
instant under consideration. 

214. Resultant Motion. If a material point receives a single im- 
pulse in any direction, it will move in that direction with a certain 
velocity. If it receives at the same instant two impulses in different 
directions, it will obey both, and move in an intermediate direction 
with a velocity differing from that of either impulse alone. The posi- 
tion of the point at the end of the instant is the same as it would 
have been had the motions, due to the impulses, occurred in successive 
instants. This would also be true for more than two motions. The 
motion which occurs as a consequence of two or more impulses is 
called the Resultant, and the separate motions, which the- impulses 
acting singly would have caused, are called the Components. 

215. Parallelogram of Motion. Suppose the point a (Fig. 279) to 
have simultaneously the two component motions represented in mag- 

c ^,j nitude and direction by ab and ac. Then the 

resultant is ad, the diagonal of the parallelogram 
of which the component motions ab and ac are 
the sides. Conversely, the motion ad may be 
resolved into two components, one along ab, and 
the other along ac, by 

drawing the parallelogram dbdc, of which it will 

be the diagonal. 

Any two component motions can have but one 

resultant, but a given resultant motion may 

have an infinite number of pairs of components. 

In the latter case there is a definite solution 

provided the directions of both components are 

known or the magnitude and direction of one. 

If the magnitudes of both components are known there are two possible 

solutions. Thus in Fig. 280, where ad is the given resultant, if the 




Fig. 279 





226 ELEMENTS OF MECHANISM 

two components have the magnitudes represented by ac and ab, the 
directions ac and ab will solve the problem, or the directions aci and 
abi will equally well fulfil the conditions. 

216. Parallelopiped of Motions. If the three component motions 
ab, ac, and ad (Fig. 281) are combined, their resultant af will be the 
diagonal of the parallelopiped of which they are the edges. The mo- 
tions ab and ac, being in the same plane, can be combined to form the 

resultant ae; in the same way ae 
and ad can be combined, giving the 
resultant af. Conversely the motion 
af may be resolved into the com- 
ponents ab, ac, and ad. 

To find the resultant of any 

number of motions : First, combine 

any two of them and find their resultant; then, combine this resultant 

with the third, thus obtaining a new resultant, which can be combined 

with the fourth; and so on. 

217. Composition and Resolution of Velocities. If the motions 
referred to in the preceding paragraphs are assumed to be uniform and 
to take place in a unit of time, the lines in Figs. 279, 280, and 281 may 
be considered as representing the velocities as well as the motions. If 
the motion of the point is variable and the lines ac and ab (Figs. 279 
and 280), or ac, ab, and ad (Fig. 281), represent the velocities imparted 
to the point a by the impulses acting in the respective directions at 
the instant, then the fines ad (Figs. 279 and 280) and af (Fig. 281) rep- 
resent the actual, or resultant, velocities at that particular instant, but 
not the actual motions. In other words (Fig. 279), ad indicates the 
motion which a would have in a unit of time if the impulses now acting 
continued unchanged, and therefore indicates the present velocity, but 
the motion was assumed to be variable, therefore present conditions 
are instantaneous only, and ad does not indicate the actual motion but 
only the instantaneous tendency. 

218. Relation between Linear Velocities of Rigidly Connected 
Points. If two points are so connected that their distance apart is 
invariable and if their velocities are resolved into components at right 
angles to and along the straight line connecting them, the components 
along this line of connection must be equal, otherwise the distance 
between the points would change. 

This may be seen by considering the center line of the connecting 
rod of a four-bar linkage, Fig. 282. 

Assuming that the crank AB is turning at such an angular speed 
that the crank pin B has a linear velocity represented by the line Bb, 



FOUR-BAR LINKAGE 227 

let it be required to find the line which represents, at the same scale, 
the linear velocity of C. Since Bb indicates the velocity of B at the 
instant, B would, if not restrained by the crank, move to a point repre- 
sented by 6 in a unit of time. The same point would be reached if 
for a part of the unit of time B should have the velocity represented 
by Be, moving along the line CB extended, to the point e which is the 
foot of a perpendicular let fall from b to CB, then from e moving out 
along this perpendicular with a velocity represented by eb. That is, the 
velocity of B may be considered as the resultant of the component Be, 
along the connecting rod, and the component Bei (equal to eb) of rota- 
tion about some point on the center line of the connecting rod. Al- 



Fig. 282 

though B does not actually go to b its tendency at the instant is to go 
there in a unit of time, and its tendency to move along the line CB is 
such that, if there were no components at right angles to CB, it would 
go to e in a unit of time. It is this tendency of B to move along the fine 
CB, that is, the component of Bb along the line CB which causes C to 
move, and since CB is a rigid rod which can be neither lengthened nor 
shortened, C must have a component velocity (or tendency to move) 
along the line CB equal to that of B, that is, equal to Be. The direc- 
tion of the other component which, when combined with this one, will 
give the actual velocity of C must be at right angles to the rod BC, 
because this component has no effect along the rod. 

The actual direction of the velocity of C is along a line Ck perpen- 
dicular to DC. Then, to find the velocity of C lay off Cf equal to Be 
and through / draw a line perpendicular to CB meeting Ck at h. Ch 
then represents the velocity of C at the same scale at which the velocity 
of B is represented. For example, if Ch is found to be three-fourths as 
long as Bb, it will indicate that at that instant C has a velocity which 
is three-fourths of that of B. The above method really consists, then, 
of treating the whole rod as if all points were given an impulse in the 



228 



ELEMENTS OF MECHANISM 



direction CB such as to cause them to have velocities equal to Be along 
CB and, simultaneously, each point were given another impulse at right 
angles to CB, causing all points to turn about some point on the center 
line of the rod. 

219. Instantaneous Axis of Connecting Rod. The discussion in 
the preceding paragraphs shows that one end of the connecting rod in 
Fig. 283 is moving at the instant in the direction Bb, perpendicular to 
AB, while the other end is moving in the direction Ch, perpendicular 
to DC. The direction of B is the same wherever the axis of rotation 
is located on the line AB or AB extended. Similarly, the direction of 
C is the same wherever its axis of rotation is located on DC or DC 










-'T 






,-»- 




B 


„ 


-" 




!.-"' 




G 


b 










Fig. 


283a 





Fig. 283 



Fig. 283b 



extended. Hence, both these points B and C might be treated for the 
instant as if they were turning about the point which is common to AB 
and DC, that is, their point of intersection 0. B and C are points on 
the connecting rod BC, as well as on the cranks, therefore, the whole rod 
BC may be treated for the instant as if it were turning about as an 
axis. The point is the trace of a line which is called the instan- 
taneous axis of BC. The principle is the same as if the connecting 
rod were made fast to a wheel which, for the moment, is turning about 
as an axis. It follows that the velocity of C is to the velocity of B 
as OC is to OB. 

If the mechanism were such that the motions of the points B and C 
were not in the same plane, the instantaneous axis would be found as 
follows: Pass a plane through the point B perpendicular to Bb; the 
motion Bb might then be the result of a revolution of B about any axis 
in that plane. In the same manner, the motion of Ch might be the 
result of a revolution of C about any axis in the perpendicular plane 



FOUR-BAR LINKAGE 229 

through C. The points B and C, being rigidly connected, must rotate 
about one axis, which in this case will be the intersection of the two 
perpendicular planes. 

If the motions of the two points B and C are in the same plane 
and parallel, as in Figs. 283a and 283b, the perpendiculars through 
B and C coincide and the above method fails. Let Bb and Ch be the 
velocities of the points B and C respectively. To find the instantaneous 
axis draw a right line through the points b and h in each case and note 
the point where it intersects BC or BC produced. This must be the 
instantaneous axis, for from the similar triangles BbO and ChO 

Bb = OB 

Ch OC' 

that is, the velocities of B and C are proportional to the distances of 
these points from 0. 

The trace of the instantaneous axis on the plane of the drawing is 
often called the instantaneous center. 

220. Centrode. If the position of the instantaneous axis of a 
connecting rod or other moving link is found for a series of positions 
through the entire cycle of motion of the linkage, and a smooth curve 
is drawn through these different positions of the instantaneous axis, the 
curve thus found is called the centrode of the link in question.* Some 
of the properties of centrodes and the uses made of them will be taken 
up later in connection with certain special examples of four-bar 
linkages. 

221. Instantaneous Axis and Centrode of a Rolling Body. The 
instantaneous axis of a rigid body which rolls without slipping upon 
the surface of a fixed rigid body must pass through all the points in 
which the two bodies touch each other, for the points in the rolling 
body which touch the fixed body at any given instant must be at rest 
for the instant, and must, therefore, be in the instantaneous axis. As 
the instantaneous axis is a straight line, it follows that rolling surfaces 
which touch each other in more than one point must have all their 
points of contact in the same straight line in order that no slipping may 
occur between them. This property is possessed by plane, cylindrical 
and conical surfaces only; the terms cylindrical and conical being used 
in a general sense, the bases of the cylinders and cones having any figure 
as well as circles. The surface of the fixed body is the centrode of the 
moving body. 

* In reality the centrode is the locus of the instantaneous center, the locus of the 
instantaneous axis being a surface sometimes called the axoid. The term centrode 
is commonly used for the locus of the axis itself as well as for the locus of its trace. 



230 



ELEMENTS OF MECHANISM 



222. Velocities of Points on Rolling Bodies. Let B, Fig. 284, be 
the center of a wheel resting on a track at the point A. If a pull is 
exerted at the center along the line Bb, parallel to the track, B will 




move in the direction Bb. The effect is the same for the instant as if 
AB were a crank turning about A as an axis and since A B is an im- 
aginary line on the wheel, the whole wheel becomes a crank turning 
about A as an axis. 

Any other point on the wheel, as E, must at the instant in question 
be moving about A as an axis, and therefore the direction of the veloc- 
ity of E is Ee perpendicular to the line AE. The magnitude of this 
velocity is to the magnitude of the velocity of B as A E is to AB. 

This velocity Ee is made up of rotation about the center B of the 
wheel combined with a velocity parallel to the track equal to the 
velocity of B. That is, the real velocity Ee may be considered as con- 
sisting of the components Eg (which is equal and parallel to Bb) and 
Ef perpendicular to BE and of such magnitude that the figure gefE is a 
parallelogram. 

223. Typical Problems on Resolution and Composition of Veloci- 
ties. The general principles presented in the foregoing paragraphs 
furnish the basis for the determination of the absolute and relative 
linear velocities of points on a four-bar linkage or, in fact, of points on 
any moving piece or pieces. A full and clear understanding of the 
methods of solving such problems can be gained only by studying the 
actual solutions of a number of cases. The following examples have 
been chosen with this end in view. Some of the illustrations are taken 
directly from actual machines;- others are modified in order to show 
the principles more clearly; still others are hypothetical cases involving 
constructions which are likely to occur in actual cases. Some of the 
examples involve special applications of the four-bar linkage which 



FOUR-BAR LINKAGE 



231 



will be analyzed later but which are introduced here merely as examples 
in velocities. 



Example 58. In Fig. 283 let AB and DC be the center lines of the cranks of a 
four-bar linkage, A and D being the fixed axes. Let the linear velocity of the crank 
pin B be represented by the line Bb perpendicular to AB. It is required to find the 
lines which shall represent at the same scale, the linear velocities of the crank pin 
C and of any point E on the center line of the connecting rod. 

Solution. Resolve the velocity Bb into the two components Be and Br along 
CB and at right angles to CB, respectively. Make Cf equal to Be and through / 
draw a line perpendicular to CB meeting, at h, a perpendicular to DC, drawn through 
C. Then Ch represents the velocity of C. 

To find the velocity of E it is necessary to find, first, the direction in which E is 
moving at the instant. This is done by finding the instantaneous axis of CB. AB 
and DC are produced until they meet at 0, which is, therefore, the instantaneous 
axis of CB. OE is next drawn, and the direction of the velocity E is along a line 
Em perpendicular to OE. The magnitude of this velocity is found by laying off 
En equal to Be and drawing a line through n perpendicular to CB meeting Em at K. 
EK is the required line representing the velocity of E. 

Example 59. Fig. 285 shows in a diagrammatic form the crank, connecting rod 
and crosshead of a steam engine. Let Bb represent the velocity of B. It is required 




crrrUY. 



Fig. 285 



to find the velocity of the center C of the crosshead pin and of a point E on the 
center line of the connecting rod. 

Solution. Bb is resolved into components along and at right angles to CB giving 
Be as the component along CB. Make Cf equal to Be. Now C is moving along the 
line NM parallel to the guides. Therefore, the velocity of C is Ch, found by drawing 
a perpendicular to CB at / meeting NM at h. 



232 



ELEMENTS OF MECHANISM 



The instantaneous axis of CB is at the point 0, where a perpendicular to NM 
through C meets AB produced. Knowing the position of this instantaneous axis the 
velocity of E can be found as in the preceding example. 

Example 60. With data the same as for Example 59, except that the connecting 
rod is a bent bar, as shown in Fig. 286, let it be required to find the lines representing 
the linear velocity of C and E. 






Solution. Since the curved bar BEC is rigid the straight line BC may be 
substituted for it without affecting the relations of B and C. Hence this line may 
be used as the line of connection between B and C and the velocity of C may be 
found from the known velocity of B as in Example 59. To find the velocity of 
E the instantaneous axis of BEC is found at as in Example 59. E is then joined 
to and the direction of the velocity of E is along a perpendicular to OE through 
E. Join C and E and resolve the velocity Ch into components along and at right 
angles to the line CE. The component En, of the velocity of E, along CE produced 
is equal to the component Ck of the velocity of C, along CE, the other component in 
each case being at right angles to CE. Hence lay off En equal to Ck and through 
n draw a line perpendicular to En meeting Em at in. Em then represents the 
velocity of E. 

Example 61. In Fig. 287, D is a disk keyed to the shaft A, the center of the 
disk being at B. J? is a rod, one end of which embraces the pin C on the slider; 
the other end is enlarged to form the strap S and contains a cylindrical hole just 
fitting over the disk D. If the shaft A turns at such an angular speed that a linear 
velocity Bb is imparted to the center B (Bb being perpendicular to the line from B 
to the center of the shaft A), let it be required to find the linear velocity imparted 
toC. 



FOUR-BAR LINKAGE 



233 



Solution. Since the hole in S is concentric with D its center must coincide with 
B in every position which the mechanism occupies and the mechanism is the equiva- 
lent of the linkage in Fig. 285. (This relation will be shown in another way in the 
next chapter.) Hence the problem becomes the same as in Example 59. 




Fig. 287 

Example 62. In Fig. 288 A B is a crank turning about the fixed center A. BC 
is a connecting rod pivoted to the slider S at C and prolonged to E. From E 
another link EF connects to the slider T. If Bb represents the velocity of B let it 
be required to find the velocity of F (that is, of the slider T). 

Solution. The instantaneous axis of BCE is at 0, hence the direction of the mo- 
tion of E is perpendicular to OE. Resolve the velocity Bb into components along 



A (fixed) 




and at right angles to BCE and transfer the component Bb 2 , thus found, to Ee%. 
Through e 2 draw a perpendicular to Ee 2 meeting Ee at e. Ee is the velocity of E. 
Next resolve Ee into components along and perpendicular to EF, getting Ee 3 along 
EF. Make Ff 3 equal to Ee 3 and through / 3 draw a perpendicular to Ff 3 meeting 
the line of motion of F at /. Ff is the required velocity of F. 



234 ELEMENTS OF MECHANISM 

Example 63. In Fig. 289 the connecting rod of the four-bar linkage is prolonged 
in a curve to E. If Bbi represents the velocity of B let it be required to find the 
velocity of E. 

Solution No. 1. The instantaneous axis of CBE is found at 0, hence the direc- 
tion of E is perpendicular to OE. The line of connection between B and E is the 




A {fixed) 
Fig. 289 



D{fixed 



straight line BE. Resolve the velocity of B into components along and perpen- 
dicular to BE, getting Bb 2 along BE. Make Ee 2 equal to Bb 2 and through e 2 draw a 
perpendicular to Ee 2 meeting Eei at d. Then Eei represents the velocity of E. 

Solution No. 2. (Fig. 290.) Find the velocity of C (= Cci) as in previous ex- 
amples. Then Cc 3 is the component of Cci which represents turning about some 
point on BC when the component of translation in direction CB is Cd = Bb 4 . 
Bb 3 is the component of Bb, which represents turning about the same point on BC. 
These components of turning must be proportional to the distances of B and C from 
the axis about which the turning occurs. R is this axis and is found by drawing 
b 3 c 3 cutting BC at R. It should be observed that R falls at the foot of the perpen- 
dicular to BC let fall from the instantaneous axis of BC and that its component of 
turning is zero, hence its true velocity is along CB and is equal to Bb 4 . From the 
above reasoning it follows that the motions of all points on the rod CBE relative ot 
each other are the same at the instant as they would be if the rod were turning about 
R as a fixed axis. Then Ee 3 is the component of turning of E about R and is found 

by making Ee 3 perpendicular to RE and -^ = -^j, ■ Next make the component of 

translation Eei equal and parallel to Bh and complete the parallelogram of which 
Ee 4 and Ee 3 are sides. The diagonal Eei of this parallelogram represents the actual 
velocity of E and is the same line as Eei in Fig. 289. This method of solution is of 



FOUR-BAR LINKAGE 



235 



value when the instantaneous axis falls off the paper or in such a position that it 
cannot be used conveniently. 




Fig. 290 



Example 64. (Fig. 291.) The long slider S has attached to it, at the pin C, the 
link CB, the other end of which is connected to the crank AB. At E is a link EF 
connecting &iF with the link DF which is also connected to S. If Bbi represents the 
linear velocity of B let it be required to find the linear velocity of F. 

Solution. Resolve Bbi into components along and perpendicular to CB, getting 
Bbi. Make Cc 2 equal to Bb 2 and draw c 2 ci perpendicular to Cc 2 meeting the line of 
motion of C at ci. Then Cci represents the velocity of C. The velocity of D must 
be the same as that of C, therefore make Ddi equal to Cci. Resolve Ddi into com- 
ponents along and perpendicular to DF, getting Dd 2 as the component along DF. 
Make Ff 2 equal to Ddi. Then Ff 2 is the component of the (as yet unknown) veloc- 
ity of F in the direction DF when the other component is perpendicular to DF. 
Hence the line representing the velocity of F must terminate somewhere on the 
perpendicular (Jim) to Ff 2 through f 2 . (In other words Ff 2 is the projection, upon the 
fine DF produced, of the line representing the velocity of F.) Thus, by first finding 
the velocity of D, it is possible to find the component of translation along the direction 
DF of the point F when the motion of F, for the instant, is considered as consisting 
of components of translation along DF and of turning about some point on DF. 

In a similar manner may be found the component Ffz of translation in the direc- 
tion FE when the motion of F is considered as consisting of translation along FE 
and turning about some point on FE. First find the velocity of E by locating the 
instantaneous axis of CEB. The direction of E is perpendicular to OE and the 
length of the line Eei representing its velocity is found by making Ee 2 equal to Bb 2 



236 



ELEMENTS OF MECHANISM 



and drawing e^x perpendicular to Ee% meeting Ee\ at e\. Next resolve Eei into 
components along and perpendicular to FE, getting Ee 3 . Make Ff 3 equal to Ee 3 
and at/ 3 draw a line perpendicular to Ff 3 . Since Ff 3 is the component of the velocity 
of F along FE when the other component is one of turning about some point on FE 

D S d C 




A {fixed) 



Fig. 291 



the line representing the velocity of F must terminate somewhere on the perpen- 
dicular (f 3 n) to Ff 3 through / 3 . It was shown above that the line representing the 
velocity of F must terminate on fam. Therefore the end of the desired line must be 
at the intersection of f 2 m and/ 3 w, namely at/i, and Ffi represents the velocity of F. 

Example 65. In Fig. 292, A B is a crank turning about A. The crank pin 
carries a block which works in a slot whose center line is MN. This slot is in the 
T-shaped head of the slider S which moves in the guides G. HF is the center line of 
the guides and Mn is at right angles to HF. If Bb represents the velocity of B, let 
it be required to find the velocity with which S is moving in the guides and also the 
rate at which the block is slipping in the slot MN. 

Solution. The component of Bb which is parallel to HF will indicate the rate at 
which the block is causing the slot to move, and the component of Bb which is 
parallel to MN will indicate the rate at which the block is slipping in the slot. There- 
fore Bb is resolved into the two components Be and Br (or eb) parallel to HF and 

MN, respectively, giving Be as the 
b 



N 



A 
Fixed Axis 



i 



Block 
B 
H 6 



._ 



T 



velocity of S and Br (or eb) as the 
rate of slipping of the block in the 
slot. 

The correctness of the above solu- 
tion canlbe understood if it beassumed 
that the point B is actually allowed 
to go to b and that instead of going 
over the path Bb it moves first to e 
along Be parallel to HF. In order to 
go there and still remain in the slot 
it must move the center line MN along 
with it until MN passes through e. 
Then B moves up along eb parallel to MN until it reaches b. This latter motion, 
since it is parallel to the slot, does not move the slot at all, but is simply a motion of 



Fig. 292 



FOUR-BAR LINKAGE 



237 



—S A— F 

mmmmmmfy 






slipping in the slot. Of course B does not actually go to b, since it is constrained to 
move in a circle about A and the next instant the direction of the velocity of B will 
be different and the rate of slipping 
and the velocity of S will have 
changed accordingly. As has already 
been pointed out, however, the velo- 
cities and directions shown in the 
figure are the ones which obtain for 
the instant when the mechanism is in 
the position shown. 

Example 66. Fig. 293 shows a 
mechanism the same as that shown 
in Fig. 292, except that the center 
line MN is not at right angles to HF. 
The method of solution is the same as 
described in the preceding example. 

Example 67. In Fig. 294, A B is a crank turning about the fixed axis A. The 
crank pin whose center is B carries a block working in a slot in the long arm which 
turns about the fixed axis G. The center line MN of this slot passes through G. If 




Fig. 293 




Cover 



Fixed Axis G 



Fig. 294 



Bb represents the velocity of B, let it be required to find the velocity of a point H on 
the swinging arm, on the center line MN. 

Solution. The velocity of H is, of course, the result of an angular speed of the 
arm caused by the motion of the block. This problem might be solved by the 



238 



ELEMENTS OF MECHANISM 



principle of the four-bar linkage and will be so analyzed later. At this point, how- 
ever, it will be treated by means of the principles of resolution of velocities. Let it 
be assumed that the slot is covered on the back by a strip, as shown. Then a point 
Bi on this strip which is directly in line with B is turning about the center G. It is 
necessary, first of all, to find the linear velocity of B\. Bb is resolved into two com- 
ponents, Be and Br (or eb). Be is along the direction in which Bi is moving, that is, 
perpendicular to BG. eb is parallel to the center line of the slot, the figure Bebr, as 
in the previous cases, being a parallelogram. Be then represents the velocity of Bi. 
The velocity of H is to Be as GH is to GB. Since GB and H are in a straight line the 
above proportion can be made graphically by drawing from G a straight line through 
e meeting at k the perpendicular to GH through H. The triangles GBe and GHK 
are similar, therefore HK is to Be as GH is to GB. Hence HK is the required veloc- 
ity of H. 

Example 68. Fig. 295 shows a mechanism similar to that in Fig. 294, except 
that the center line of the slot does not pass through G. 

Solution. The principles involved in the solution are the same as those described 
for Example 67. The component Be is perpendicular J;o GB and not to MN. H 
\k 




Fixed 
Axis A 



Fig. 295 




has a velocity perpendicular to GH and bearing the same ratio to Be that GH bears 
to GB. The similar triangles can be constructed by laying off on GH a distance 
GBi equal to GB, drawing B 2 L perpendicular to GB 2 and equal in length to Be, 
then drawing from G through L to meet the perpendicular to GH through H at K. 
HK is then the required velocity of H. 



FOUR-BAR LINKAGE 



239 



Example 69. The mechanism in Fig. 296 is the same as that in Fig. 295, except 
that the slot is curved instead of straight. The method of solution is the same. The 
only new point to be noticed is that the line Br is tangent to the center line of the 
slot at B 




A (Fixed) 



Fig. 297 



Example 70. In Fig. 297 the curved bar BR slides through the block which is 
pivoted at C on the crank DC. The lower end of the bar is carried by the crank 
AB. C represents the axis of the pin by which the block is attached to the crank 
DC, while Cx represents the point on the bar BR which is inside the block where the 
axis of the pin intersects the center line of the bar. If Cci represents the linear 



240 ELEMENTS OF MECHANISM 

velocity of C and Bbi the velocity of the axis of the pin B, let it be required to find 
the linear velocity of a point R on the rod. 

Solution No. 1. (Fig. 297.) The component of the velocity of Cx normal to 
the sliding surfaces, when paired with another component in the direction of sliding, 
must be equal to the component of Cci normal to the slide when paired with another 
component along the slide. This component normal to the slide is C x c$. Also, 
the component C x c 2 along the line joining C x and B, when paired with another com- 
ponent perpendicular to C x c 2 , is equal to Bb 2 . Hence the velocity of C x is C x Ci and 
the instantaneous axis of BC X R is at 0. The direction of R is therefore perpen- 
dicular to OR and the length of the line Rr ± which represents the velocity of R is 

found from the equation j~ = ^~-- 

Solution No. 2. (Fig. 298.) First assume that the axis c of the block is fixed 
(that is, Cci = 0). Then the instantaneous axis of BR would be at Oi and the line 
representing the velocity of R would be represented by Rr 6 perpendicular to OiR 

and of a length such that -^ = 7^-5 

Next assume B to be fixed (that is, Bbi = 0). Then the velocity of C x would 
be represented by C x c & , found as in Example 68, and the velocity of R would be 

represented by Rn perpendicular to BR and of a length such that -—■ = -^- • 

The line representing the actual velocity of R is Rr x which is the diagonal of the 
parallelogram of which Rr 6 and Rr 6 are sides. 

Example 71. In the mechanism shown in Fig. 299, a wheel turning about the 
axis D has pinned to it at C a link connected to the end of the crank AB. The 
wheel also carries a pin E on which is a block. Through the block slides the rod 
RHF. The point H is connected to the pin B and the end F is attached to the 
slider. The wheel is turning at such an angular speed that the axes of the pins 
C and E have velocities represented by Cci and Ee\ respectively. Let it be required 
to find the line representing the velocity of any point as R on the rod RHF. 

Solution. First find the velocity of B and resolve this velocity Bh into com- 
ponents along and perpendicular to BH, thus getting Bb s . The instantaneous axis 
of RHF lies somewhere on the perpendicular to XX through F but its exact position 
on this line is not yet known. Hence the velocity of H cannot be found as yet, 
H is a point which is common to BH and RHF and the component of its velocity 
along BH must be equal to Bb 3 . Therefore the component along BH of the velocity 
of any point on RHF which lies along BH or BH produced must also equal Bb. 
(it being understood that the other component is perpendicular to BH). If RHF 
is assumed to be enlarged to include the point K where BH produced intersects the 
perpendicular to XX through F, then K, being a point on RHF lying along BH S 
will have the component Kk 3 equal to Bb 3 . The actual velocity of K must be per- 
pendicular to FK, hence Kki is the velocity of K. Let E x represent the point on 
the axis of RHF where it intersects the axis of the pin E. Then resolve Kh into 
components perpendicular to and along KE X getting Kki. Make E x e± equal to 
Kki and draw a perpendicular to E x e t through e 4 . Next resolve the velocity Ed 
into components along and normal to the sliding, getting Ee 2 as the normal compon- 
ent. The normal component of the velocity of E x must be the same as Ee 2 when 
paired with another component in the direction of the sliding. Hence the line 
Exe 3 drawn to the intersection of e 2 ei and e 4 e 3 represents the actual velocity of E x . 



FOUR-BAR LINKAGE 



24i 



The instantaneous axis of RHF then is found at 0. The velocity of R is therefore 



represented by Rn, perpendicular to OR and of such length that 



Rn OR 
Kh ~ OK 




Fig. 298 

Example 72. Rate of Sliding of Gear Teeth. To determine the rate of sliding of 
one gear tooth upon another at any position it will be necessary to find the linear 
velocity of the point of contact D (Fig. 300) in each of the teeth, and resolve these 
linear velocities into their components along the common normal and the common 
tangent. Let DE represent the linear velocity of D around Oij the components 
of DE along the common normal and tangent are DF and DG respectively. The 



242 



ELEMENTS OF MECHANISM 




Fig. 299 



FOUR-BAR LINKAGE 



243 



direction of the motion of D around 0% is along the line DH. To find the magnitude 
of its linear velocity we have DF as its component along the common normal, since 
this normal is the line of connection between the two sliding surfaces, and com- 




Fig. 300 



ponents along the line of connection must be equal. This will give DH as the linear 
velocity of D around O2, and DK as its component along the common tangent. The 
rate of sliding will be found to be GK , equal to DG + DK, since the components along 
the tangent act in opposite directions. 



CHAPTER XI 
LINKWORK 

224. A Link may be defined as a rigid piece or a non-elastic sub- 
stance which serves to transmit force from one piece to another or to 
cause or control motion. 

For example, that part of a belt or chain running from the driven to 
the driving wheel, the connecting rod of an engine, the fluid (if assumed 
to be incompressible) in the cylinder of a hydraulic press, would be 
links according to the above definition. In ordinary practice, how- 
ever, the name is applied to a rigid connector (see § 7), which may 
be fixed or in motion. 

225. A Linkage consists of a number of pairs of elements con- 
nected by links. If the combination is such that relative motion of 
the links is possible, and the motion of each piece relative to the others 
is definite, the linkage becomes a Kinematic Chain. If one or more of 
the links in such a chain be fixed the chain becomes a Mechanism 
(see § 2). 

In order that a linkage may constitute a kinematic chain, the num- 
ber of fixed points or points whose motions are determined by means 
outside the particular linkage in question, must bear such a relation 
to the number of links in the linkage that the linkage may form a 
four-bar linkage or a combination of two or more four-bar linkages. (See 
§ 210.) This may be seen by reference to Figs. 301, 277 and 302. 




Fig. 301 

The linkage in Fig. 301 consists of three finks AB, AC, and BC, form- 
ing a triangle and it is apparent that no relative motion of the links 
can occur since only one triangle can be formed from three given lines. 

244 



LINKWORK 245 

On the other hand, if four links are involved, as in Fig. 277, relative 
motion of a definite nature will result. If now five links, as AB, BC, 
CD, DE, and EA, Fig. 302, constitute the linkage, any link, as AE, may 
be fixed; then AB and ED become cranks, but a given angular motion 
of the crank AB does not impart a definite resulting angular motion to 



Fig. 302 

DE unless the point C is guided by some external means. If, how- 
ever, C is guided by the crank FC turning about the fixed center F the 
motions of all the links become determinate. But the linkage, by the 
addition of the crank FC, has now been transformed into a combination 
of two four-bar linkages, namely: ABCF and FCDE with A, F, and E 
fixed. 

In general, it may be said that any mechanism may be analyzed as a 
four-bar linkage or as a combination of two or more such linkages. 

226. The Four-Bar Linkage. In § 210 a four-bar linkage was de- 
scribed and special names were given to the links when one of them was 
fixed. It is not necessary that any link be absolutely at rest. The 
fink AD, Fig. 277, which is there assumed to be fixed, may be attached 
to some other part of the machine which itself is in motion. ABCD 
remains a four-bar linkage and the relative motions of its four links are 
unchanged although, of course, the absolute motion of each fink de- 
pends not only upon its motion with relation to the link (in this case 
AD) which is assumed to be fixed, but also upon the motion which 
that link has. 

In view of that which has preceded, a four-bar linkage may be de- 
scribed as consisting of four cylindrical pairs of elements, or their equiv- 
alent, each element of each pair being connected by a rigid link, or its 
equivalent, to one of the elements of an adjacent pair. Thus, the pin 
A, Fig. 272, is connected by the fixed frame E to the pin D. The en- 
larged end of the crank F which contains the cylindrical hole paired 
with A is connected to one of the pair at C, and so on. 



246 ELEMENTS OF MECHANISM 

227. Relative Motion of the Links in a Four-Bar Linkage. Since, 
as shown in § 226, the motions of the links, relative to some one link 
assumed to be fixed, are not changed if motion is imparted to that link, 
it follows that the motion of any link, relative to any other link of the 
linkage, is the same whichever link is fixed. In other words, the rela- 
tive motions of the links of a four-bar linkage are independent of the fixed- 
ness of the links. This principle is taken advantage of in the applica- 
tion of four-bar linkages, particularly in cases where centrodes are 
substituted for some of the links, as will be illustrated later. 

The laws relating to the motions of the links can be studied more 
conveniently by assuming one link fixed, and this method will be 
followed in the succeeding paragraphs. 

228. Angular Speed Ratio of Cranks. The angular speeds of the 
two cranks of a four-bar linkage are inversely as the lengths of the perpen- 
diculars or any two parallel lines drawn from the fixed centers to the center 
line of the connecting rod; also, inversely as the distances from the fixed 
centers to the point of intersection of the center line of the connecting rod 
and the line of centers (produced if necessary) . 

This law may be shown to be true in two ways. 
1°. By reference to the instantaneous axis of the connecting rod. 
Let ABCD (Fig. 303) represent the linkage, AD being the fixed link. 
To show that 

Angular speed of DC _ Am _ Ak _ 

Angular speed of AB Dn Dk 




The instantaneous axis of BC is at and since the linear speeds of 
B and C are to each other as their distances from the instantaneous 

axis ''see § 219) 

Linear speed C _ OC „. 

Linear speed B OB 



LINKWORK 247 

But since the angular speed of DC is equal to the linear speed of C di- 
vided by the radius DC (see § 38^ it follows that 



A , , -p.,- Linear speed C 
Angular speed DC = jjn 



(II) 



Similarly- 
Dividing (II) by (III) 



Angular speed AB = Line ^speed ff (III ) 



Linear speed C 

Angular speed DC _ DC Linear speed C AB . . 

Angular speed AB Linear speed B Linear speed B DC 
AB 

Whence, combining (I) and (IV), 

Angular speed DC ^OC AB PC AB 

Angular speed AB OB X DC DC OB ^ V) 

Draw Ot, Dn and Am perpendicular to BC. Then, from the similar 
triangles OtC and DnC, 

oi = qc^ 

Dn DC' 



a- .-., , Am 
Similarly -~7 = 


AB 
= OB 




Substituting these values in (V) 






Angular speed DC 
Angular speed AB 


Ot Am 
~~ Dn X Ot'' 


Am 
~ ~Dn 



(73) 

Producing CB to meet DA at k, it will be noticed that triangles 
Dnh and Amk are similar, hence 

Ak _ Am 
Dk~ Dn' 



Therefore, from equation (73), 

Angular speed DC _ Ak^ 
Angular speed AB Dk 



(74) 



In determining the speed ratio of the cranks for any specific case 
either equation (73) or (74) may be used as happens to be more con- 
venient. 



248 ELEMENTS OF MECHANISM 

2°. By resolution of velocities. In Fig. 304 let Bbi represent the 
linear velocity of B. Then Bb 2 is the component of this velocity along 
CB and Cci is the velocity of C. 

QA 

\ 




Fig. 304 

Therefore Linear speed C = Cci j 

Linear speed B Bbi 
and since linear speed is equal to angular speed divided by the radius, 
and the triangles Ccic 2 and DCn are similar as are also Bbib 2 and ABm, 
this may be written 

Angular speed DC _ Cci . Bbi _ Cc^ Am 
Angular speed AB~DC^AB~Dn X Bfo' 

But Cc 2 = Bb 2 . 

Hence the equation becomes 

Angular speed DC _ Am 
Angular speed AB Dn 
as in (73). 

This angular speed ratio of course varies for every relative position 
of the links; but if the perpendicular from the instantaneous axis to 
the center line of the connecting rod should fall at the intersection of 
the center line of the connecting rod and the line of centers, that is, in 
Fig. 303, if the points t and k should coincide, the angular speed ratio 
is essentially constant for slight movements in either direction. The 
same would be true should the points B and C be moving in lines par- 
allel to each other. 

229. Diagrams for Representing Changes in the Linear Speed 
Ratio or Angular Speed Ratio in Any Linkage. To obtain a clear 
knowledge of the change in velocity ratio in any linkage a diagram may 
be drawn where the abscissae may represent successive positions of one 
of the oscillating links, and the ordinates represent the angular speed 
ratio of the oscillating links. A smooth curve through the points thus 
found would show clearly the fluctuations in the angular speed of one of 
the links relative to the other. A curve for linear speed ratio could be 
similarly plotted. 



LINKWORK 



249 



In the linkage shown in Fig. 305 let AB turn uniformly left handed; re- 
quired a curve to represent the ratio — — -. — =—:-= for a complete 

M angular speed AB 

rotation of AB. Take positions of A B at intervals of 30° and draw per- 



\ P* A 




Fig. 305 



0; 



pendiculars from D and A to BC in each of its positions. The ratio of the 
two perpendiculars in each position will give the angular speed ratio : 
thus, starting with AB as 
given in the figure, 

angular speed DC _ 

angular speed AB 
in the position ABiCiD 
we have 

angular speed DC _Anh % 
angular speed A B Dn\ 
etc. Plotting these values 
as ordinates and the 30° 
positions of'AB as ab- 
scissae will give the curve 
shown in Fig. 306. Ordi- 
nates above the zero line 
indicate that the cranks 





/ 


S~~ 




















/ 


/ 






















/ 










\ 














/ 










\ 














3 


U 6 


0° 9 


fr I'* 


0°li 


0° 11 


yy j 2 


0°2a 


U 2, 


0°3C 


J°3. 


0°3( 
















































1 


Ord 


inat< 


18-- 


Angul 


arS/- 


eed L 


re. 












Angular Sf 


:-€J , 


.8, 


Abs 


dss 


ie - Position i 
f 1 


" 


A.B, 













Fig. 306 



turn in the same direction, while those below show that the directions 
of rotation are opposite. 

230. Dead Points. A position in the cycle of motion of the driven 
crank of a linkage in which it is in line with the connecting rod, and 



250 



ELEMENTS OF MECHANISM 



therefore cannot be moved by the connecting rod alone, is known as 
a dead point. If the driven crank makes complete revolutions there 
are two such positions in its cycle. 

231. Crank and Rocker. Let the link AD (Fig. 307) be fixed, and 
suppose the crank AB to revolve while the lever DC oscillates about 
its axis D. In order that this may occur, the following conditions must 
exist. 



1° AB + BC + DOAD, 3 C 

2° AB+'AD\+DC>AB, 4 C 



AB + BC - DC < AD, 
BC -AB + DOAD. 



1° and 2° must hold in order that any motion shall be possible; 
3° can be seen from the triangle AC 2D in the extreme right position 




B*\ 



Fig, 307 



AB2C2D, which must not become a straight line; and 4° can be seen 
from the triangle AC J), in the left extreme position AB\C\D. 

There are two points C\ and C2 in the path of C at which the motion of 
the lever is reversed, and it will be noticed that if the lever DC is the 
driver, it cannot, unaided, drive the crank AB, as a pull or a thrust on 
the rod BC would only cause pressure on A, when C is at either d. or 
C2. If AB is the driver, this is not the case. 

The above form of linkage is applied in the beam engine as shown 
in Fig. 308, the fixed link AD being formed by the engine frame; corre- 
sponding parts are lettered the same as in Fig. 307. The instantaneous 
axis is, for the position shown, at O, and for the instant the linear speed 
of B is to the linear speed of C as OB is to OC, or as Bf is to Ce, the 
line ef, drawn parallel to BC, being made use of when the point O comes 
beyond the limits of the drawing. 

The angle through which the lever DC (Fig. 307) swings can be 
calculated for known values of AB, BC, CD, and DA. 

From the triangle AC 2D 

DC 2 + AD 2 - (BC + AB) 2 



COS ADC2 



2 (DC X AD) 



LINKWORK 



251 



and from the triangle AC J) 

cos ADC, - DC 2 + AD 2 - (BC - AB)> 

C0S ADCl ~ 2 (DC X AD) 

and C1DC2 = ADC 2 - ADd. (75) 

Thus the two angles ADC 2 and ADCi can be calculated, and their 
difference will give the angle required. 

-,0 




Fig. 308 

If tne link BC is made stationary, the mechanism is similar, the only- 
difference being in the relative lengths of the connecting rod and 
stationary piece or frame. 




Fig. 309 

Fig. 309 shows a case of the crank and rocker as applied in wool- 
combing machinery. Here the crank AB turns uniformly on its axis A, 
while CD oscillates about D ; both axes A and D are attached to the 



252 ELEMENTS OF MECHANISM 

frame of the machine, which forms the fixed link AD. The connecting 
rod CB is prolonged beyond B, and carries a comb E at its extremity, 
which takes a tuft of wool from the comb F and transfers it to the comb 
G, both combs F and G being attached to the frame of the machine. 
The full lines show the position of the links when the comb E is in the 
act of rising through the wool on F, thus detaching it, and the dotted 
lines show the position of the links when the comb E\ is about to deposit 
the tuft of wool on G. The same combination inverted is used in some 
forms of wool-washing machines. 

232. Drag Link. If the link AB, in the linkage Fig. 307, is 
made the stationary piece or frame, as in Fig. 310, the links BC and 



Fig. 310 

AD revolve about B and A respectively, that is, become cranks, and 
CD becomes a connecting rod. This mechanism is known as the drag- 
link. 

In order that the cranks may make complete revolutions, and that 
there may be no dead-points, the following conditions must hold: 

1° Each crank must be longer than the line of centers, which needs 
no explanation. 

2° The link CD must be greater than the lesser segment C 4 / and less 
than the greater segment C 4 Z>2, into which the diameter of the greater 
of the two crank circles is divided by the smaller circle. This may be 
expressed as follows : 

CD > AB + AD - BC (see triangle AdDJ, 
CD< AD + BC - AB (see triangle BC 2 D 2 ). 



LINKWORK 



253 



Producing the center line of the connecting rod until it intersects the 
line of centers at e, and dropping the perpendiculars Am and Bn upon it, 

Angular speed AD _ Be _ Bn 
Angular speed BC Ae Am 

In the positions ABC1D1 and ABC 3 D 3 , when CD is parallel to the 
line of centers, the angular speeds of AD and BC are equal, since the 
perpendiculars Bn and Am then become equal. 

If BC revolves left-handed and is considered the driver, it will be 
noticed that between the positions ABC 3 D 3 and ABC1D1 the crank AD 
is gaining on BC, and between ABC1D1 and ABCsD 3 it is falling behind 
BC. 

Fig. 311 shows an application of the drag link for driving the ram of 
a Dill slotter. The links in this figure are lettered to correspond with 
Fig. 310. The large gear, turning on a fixed boss, centered at B on the 




Fig. 311 

frame, carries the pin C and forms the driving crank corresponding to 
BC, Fig. 310. The shaft A has its bearing in a hole in the large boss 
on which the gear turns and has keyed to it the crank arm AD. On 
the other end of this shaft is another crank arm, or its equivalent, the 
center line of which is AN. To this latter crank arm is attached 
the connecting rod which drives the ram. The mechanism is shown 



254 



ELEMENTS OF MECHANISM 



in the position which it occupies when the ram is about at the middle 
of the downward or cutting stroke. 

233. The Double Rocking Lever (Fig. 312) shows the same linkage 
with DC as the fixed link. In this case the cranks CB and DA merely 

oscillate on their axes C and 
D. The extreme positions 
may be assumed at AiBi and 
A 2 B 2 . 

234. Parallel Crank Four- 
Bar Linkage. In Fig. 313, 
the crank AB is equal in 
length to the crank CD and 
the line of centers AD is 
equal to the connecting rod 
BC. The center lines of the 
linkage thus form a parallelo- 
gram in every position, provided the cranks turn in the same direction. 
Therefore, the perpendiculars Am and Dn are always equal and the 
two cranks are always turning at the same angular speed. A familiar 
example of this linkage is furnished by 
the cranks and parallel rod of a loco- 
motive. In this case the link formed 
by the center line of bearings in the 
frame carrying the axles of the two 
driving wheels corresponds to the line 
of centers, but is itself in motion. 

235. Non-Parallel Equal Crank Linkage. 




Fixed Axis 



Fixed Axis 



Fxg. 313 



In the linkage shown in 

Fig. 314, AB is equal to CD and AD is equal to BC. Provision is made, 

ihowever, to cause the cranks to turn in opposite directions; in which 

case the perpendiculars Am and Dn do not remain equal to each other. 

Therefore, if the crank AB turns with 

uniform angular speed, the crank DC has 

a varying angular speed, although both 

make one complete turn in the same 

length of time. 

The opposite directions of revolution 
may be secured by providing some 
means of causing the cranks to pass the dead points in the proper 
direction. This may be accomplished by means of some device placed 
at the instantaneous axis of the connecting rod when in these positions 
although the entire linkage may be replaced by gears whose pitch sur- 
faces are the centrodes of two of the opposite links. If in the linkage 




LINKWORK 



255 



in Fig. 314, AD being the line of centers, the centrode of BC is drawn 
it will be found to be the hyperbola, part of which is shown in full lines 
in Fig. 315, with foci at A and D and transverse axis fg lying along the 




fixed link AD and equal in length to AB = DC. Similarly, if BC is 
considered as the fine of centers the centrode of AD is the hyperbola, 
part of which is shown in dotted lines in Fig. 315, with foci at B and C 




Fig. 316 "o 

and transverse axis hk lying along BC and equal in length to AB = DC. 
If, with A and D as the fixed centers, a pin is placed on the connecting 
rod BC at h (Fig. 316) and a corresponding open eye at g on the fixed 



256 ELEMENTS OF MECHANISM 

line AD, the points h and g being the same points as h and g in Fig. 315, 
the pin at h will mesh into the eye at g when the linkage reaches one 
dead point and thus insure that the cranks pass the dead point in the 
proper direction. A similar pin at k and eye at / will provide for 
passing the other dead point. 

If the link DC is assumed to be the fixed link the centrode of AB is 
the ellipse N, Fig. 317, with foci at D and C and the length of the major 




Fig. 317 

axis is equal to AD= BC. Similarly if AB is assumed to be fixed the 
centrode of DC is the ellipse M with A and B as foci and with AD = BC 
as the length of the major axis. 

The ellipses being closed curves, it is possible to substitute for the 
links, short cylinders whose bases are of the form of the elliptical cen- 
trodes. Let it be supposed that two equal gears are made whose 
pitch surfaces are of the form of the ellipses shown in Fig. 317, 
and that these pitch ellipses are placed in contact with each other as 
in the figure. Links BC and DA may be attached to the gears at 
the foci of the ellipses, and if the gear N is held still and M rolled 
around it the links CB and DA will turnabout the centers C and 
D respectively and the line joining the foci A and B will have the same 
motion that it would have in the linkage. Or M may be pivoted at the 
focus A and N at the focus D and both allowed to turn, the result being 
the same as in the linkage with AD as the fixed link, AB and DC as 
cranks. 

If the cranks are prolonged as shown in Fig. 318 and pins placed at 
n and mi with corresponding eyes at m and Wi, these points being the 
same as the points with corresponding letters in Fig. 317, a means 
is provided for passing the dead points if the links themselves are 
used. 

If desired, one dead point may be passed by means of pin and eye on 
the elliptical centrodes and the other dead point by pin and eye on the 
hyperbolic centrodes. Or a few teeth on sectors of gears at these parts 
of the centrodes might be used in place of the pin and eye. 



LINKWORK 



257 



Elliptic gears designed on the above principles have been used to 
drive the rams of machine tools, such as slotters, so as to give a slow 
cutting stroke to the tool, and a quicker return stroke. In applying 




Fig. 318 

the gears for such a purpose one of them, as, for instance, M, is on a 
shaft at A driven at a uniform speed from some external source of 
power. The other gear N is on a shaft at D to which is attached the 
crank or other device for moving the ram. 

236. Slow Motion by Linkwork. The four-bar linkage can, if 
properly proportioned, be made to produce a slow motion of one of the 
cranks. Such a combination is shown in Fig. 319, where two cranks 




Fig. 319 



AB and DC are arranged to turn on fixed centers and are connected 
by the link BC. If the crank AB is turned right-handed, the crank 
DC will also turn right-handed, but with decreasing speed, which will 
become zero when the crank AB reaches position AB X in line with the 
link BC\ any further motion of AB will cause the link DC to return 



258 



ELEMENTS OF MECHANISM 



toward its first position, its motion being slow at first and then gradually- 
increasing. This type of motion is used in the Corliss valve gear, as 
shown in Fig. 320. The linkage ABCD, moving one of the exhaust 




Fig. 320 

valves, will give to the crank DC a very slow motion, when C is near &, 
when the valve is closed, while between C and C 2 , when the valve is 
opening or closing, the motion is much faster. The same is true for 
the admission valves, as shown by the linkage AEFG. 

237. Linkwork with a Sliding Pair. In Chapter X (§211) there 
was shown the relation between a linkage, such as the one in Fig. 321, 
and the simple four-bar linkage. It is important that this relation be 
clearly understood before proceeding and it is suggested that the reader 
review that discussion before studying what follows. 




At Infinity 

Fig. 321 

Referring now to Fig. 321, the four links of the linkage are AB,.BC, 
CD and AD, the lines AD and CD meeting at infinity, that is, being 
parallel, and perpendicular to the center line of the slot in T. It should 
be borne in mind that the piece T is not one of the links of the four-bar 



LINKWORK 



259 



linkage, but that T and the block pivoted to the link BC, replace, and 
constitute the equivalent of, the two infinite links AD and CD. 

Four distinct mechanisms may be obtained from this form of the 
linkage by making each one of the links, in turn, the stationary link. 
These four mechanisms, with applications, will now be discussed. 

238. The Sliding-Block Linkage. Considering the piece T (Fig. 
321) as fixed gives the mechanism shown in Fig. 322 which is the 




{Fixed Axis) j 



Li 



Fig. 322 

mechanism commonly used in pumps and direct-acting steam engines. 
When employed in a steam engine, the block at C, called the cross- 
head, is the driver and the crank A B the follower; in a pump the 
reverse is the case. 




Fig. 323 

Movement of Crosshead. In Fig. 323 let AB represent the crank, BC 
the connecting rod, and mn the path of the point C in the crosshead. 



260 



ELEMENTS OF MECHANISM 



The travel of the crosshead mn is equal to twice the length of the 
crank AB and the distance of C from A varies between BC + AB 
= An and BC — AB = Am, AB being the length of the crank, and 
BC the length of the connecting rod. 

To find the distance the point C has moved from n, the beginning 
of its stroke or travel, let the angle made by the crank with the line 
An be represented by 8, and draw Bg perpendicular to An. The move- 
ment of the crosshead from the beginning of its stroke is, for the angu- 
lar motion 6 of the crank, 

Cn = An — AC = An — {Ag + gc). 
From the right triangle BCg 

VW 2 

Hence 



gC = VBC-Bg\ 

Cn = An- AB cos d - ^W 

= AB + BC-AB cos d - VW 



AB sin 2 d 



AB (1 - cos 6) + BC \ 1 



AB' sin 2 d, 
v/l-St'sin^l 



BC 



(76) 
(77) 



It will be noticed that Equation (77) indicates that the displacement 
differs from that which C would have if its motion were harmonic 
(assuming AB to turn with uniform speed) by the term 



BC 1 



vT 



Alf 
BC 2 



(See Equation 5, page 7.) 

and that the value of this term decreases as BC increases relative to AB. 

That is, the longer the connecting rod is made relative to the crank the 

more nearly the motion of the cross- 
head approaches harmonic motion. 
The motion may be represented 
graphically by plotting a curve, 
where the ordinates represent suc- 
cessive values of Cn, and the abscissae 
represent angular positions of the 
crank AB. Fig. 324 shows the curve 
for the linkage given in Fig. 328, the 
dotted line being the corresponding 
curve, if the motion of C were har- 
monic. 
If a device similar to that shown in Fig. 325 is used, in which a rigid 

rod is attached to the crosshead, this rod having a slot in it at right 




LINKWORK 



261 



angles to the line of motion of the crosshead and embracing a block 
pivoted on the crank pin, then the crosshead would have harmonic 
motion if the crank turned at uniform speed. 



-4- 



' \G ' _ _ Connec ting Rod of 



F 



-To— Cross-he a d- 



Fig. 325 

This device is the equivalent of a connecting rod of infinite length, 
and the whole mechanism may be thought of as a four-bar linkage in 
which only one of the links, namely, the crank AB, is a finite quantity. 
The line of centers and the infinite connecting rod are indicated in the 
figure. The other crank is an imaginary fine parallel to the line of 
centers at an infinite distance away. 




Crosshead Pin 
C 



Fig. 326 



Linear Speed of Crosshead. It is convenient to be able to determine 
the velocity of the crosshead, and hence of the piston, of a steam engine 
for different positions of the stroke when the speed of the crank pin is 
known. In Fig. 326 

Linear speed of C _ OC 



Linear speed of B OB 



(78) 



262 ELEMENTS OF MECHANISM 

Through A draw a line perpendicular to the center line of the slot, and 
extend the center line of the connecting rod to cut this line at m. Then 
the triangles OCB and mBA are similar. Hence, 

OC = Am 
OB AB ' 

Substituting this in Eq. (78) gives 

Linear speed of C _ Am^ . . 

Linear speed of B AB' 

or in words, The linear speed of the crosshead or piston of a steam engine 
is to the linear speed of the crank pin as the distance between the crank 
shaft and the point where the connecting rod cuts the perpendicular through 
the center of the crank shaft is to the length of the crank. 
From the similar triangles CAm and CgB 

Am AC A D AC ^Ag + gC 

-5- = -77 or Am = gB— FT = gB . ' y • 
gB gC y aC gC 

Whence 

= AB sin 9 {AB cos 6 + ^BC 2 - AB 2 sin 2 d) 
VbC 2 - AB 2 sin 2 d 
Substituting this value in (79) gives 

Linear speed of C _ . _ , AB sin d cos d . . 

Linear speed of B ~ Sm V~BC 2 - AB 2 sin* d' 

This same result may be derived by another method. Let v repre- 
sent the speed of the crosshead, s its displacement, and t the time dur- 
ing which the displacement has taken place. 

d? 
Then v = ^- (See Equation (2), § 25.) 

Letting w represent the angular speed of A B in radians per unit of 
time and expressing 6 as cot, Equation (76) may be written 

s = AB + BC - AB cos ut -VbC 2 - AB 2 sin 2 at. 
Therefore 

, » ~ ds An . _, . wAB 2 sin oot cos ut /r> _ 
Linear speed of C ■ = -r ± = coAB sm oot + — / • (81) 

dt VBC 2 -AB 2 sm*ut 

But the linear speed of B = oo AB. 

™ ,. Linear speed of C . '. AB sin oot cos oot /r) _ x 

Therefore =n — , ■ p = sm cot -\ . • (82) 

Linear speed of B V BC 2 - AB 2 sitf ut 



LINKWORK 



263 



When = 90°, Am = AB and the speeds of C and B are equal. To 
find other values of 0, when C and B have equal speeds; from Equa- 
tion (80), 

., . . . AB sin cos 
1 = sin + 



Solving this for sin gives 
Sin0 



VbC 2 - AB 2 sin 2 



4AB* 



-5C±V8A£ 2 +£C 2 



(83) 



The linear speed ratio between C and B may be shown graphically, 
using coordinate axes, the ordinates representing the ratio and the ab- 
scissae representing angular positions of the crank AB. Fig. 327 shows 
the curve for the linkage given in Fig. 328. 



Y 
1.0 




/' 


^ 


















>^- 


"x 






0.5 
0.4 
1 

0.1 


-// 


' 












\ 


X 


u i 


"' 30° 


60° 75° 90° 120° 150° 180° 










F 


IG. 


327 









Fig. 328 illustrates other methods of showing the linear speed ratio. 
In this figure the constant linear speed of B is represented by the crank 
length AB. From Equation (79) 

Linear speed C _ Am 

Linear speed B AB 

Therefore, by laying off on the line AB, which shows the crank posi- 
tion, the distance At = Am, and repeating this construction for a 




sufficient number of crank positions, the full curve AtA will be obtained 
where the intercept At on the crank fine shows the velocity of C, AB 
being the constant velocity of B. A similar curve would be found for 



264 



ELEMENTS OF MECHANISM 



the crank positions below the line MA. Similarly the full curve Nt\M 
might be obtained by laying off on the successive perpendiculars drawn 
through the point C the corresponding distances Am. The dotted curves 
are the corresponding curves for harmonic motion and are circular. 

If, in Fig. 326, the crank AB is the driver, it can always produce 
reciprocating motion in the block at C; but if the block is the driver, it 
cannot produce continuous circular motion unless some means of pass- 
ing the dead points be employed. This is usually accomplished in 
steam engines by attaching to the crank shaft a heavy fly wheel, the 
momentum of which carries the crank by the dead points. The im- 
possibility of starting at the dead points still remains. 

To obviate this difficulty two crank and connecting-rod mechanisms 
may be combined, as shown in Fig. 329, where the cranks are placed 



A, 



B : x 4 



i 



C^¥S^^§ 



□ 



Fig. 329 

at right angles to each other and joined by a shaft. This combination 
is employed in locomotives, and in hoisting and marine engines, one 
crank being very near its best position to be acted on by the rod while 
the other is at a dead point. 

Fig. 330 shows another method of passing the dead points sometimes 
used in marine engines. Here the two connecting rods BC and BCi 
are located in parallel planes and act upon the same crank AB. By 
suitably forming the ends of the rods, they might be located in the 
same plane. 

Acceleration of Crosshead. Since (from Equation (4)) a = -57* where 

a represents the acceleration, v the linear speed and t the time, Equation 
(81) may be differentiated giving 

coAB 2 sin 



a = ltV 



AB sin cot + 



Vbc 2 



cot cos cot \ 
AB 2 sin 2 at' 



LINKWORK 



265 



= co 2 AB cos cot + 



rfAB 2 (cos 2 cot 



sin 2 cot) , o) 2 AB 4 sin 2 cot cos 2 cot ,_„ 
+•7=^ — ==^r— — 73- (84) 



The acceleration may be found with a fair degree of accuracy graphi- 
cally by plotting a velocity curve as shown in Fig. 331 in which the 
abscissae are linear distances representing time (for example, each space 
may represent one second) and the ordinates are the corresponding 




Fig. 330 

velocities of C. Then if a tangent be drawn at any point as k and this 
tangent is made the hypotenuse of a right triangle kpd whose base is 
parallel to xx and equal to the distance which represents one time 




Fig. 331 

unit, then the altitude pd represents the acceleration, at the same scale 
at which the velocity ordinates were plotted. In other words, the 
slope of the tangent to the velocity-time curve represents the accel- 
eration. 



266 



ELEMENTS OF MECHANISM 



/ Fixed Axis \ 



Crank 




Fig. 332 



Fixed Axis 




Line of Centers 



Crank---" 



Fig. 333 



LINKWORK 267 

239. Swinging or Rocking Block Four-Bar Linkage. Referring to 
Fig. 321, if, instead of considering the piece T as stationary, giving the 
four-bar linkage already discussed, the piece BC is made stationary, 
the linkage becomes that shown in Fig. 332. Here BA is a crank but 
A is now the crank pin. The pin C is a fixed axis with the block swing- 
ing on it. As the crank BA revolves the piece T oscillates and, at the 
same time, slides back and forth over the block. 

This mechanism is applied in a modified form as a quick return 
motion in certain machine tools, particularly shapers. Fig. 333 is the 
same as Fig. 332, drawn with the line BC vertical and with the piece T 
extended with a connection at N to drive the ram of the shaper. The 
apparent objection to using the mechanism in this form is that the point 
N moves up and down atfthe same time that it swings. The only real 
purpose of the sliding of the piece T over the block is to allow the dis- 
tance between A and C to change as the crank BA revolves. This is 
accomplished equally well if the block is pivoted on the pin A and the 
arm T swings on a pin at C, as shown in Fig. 334. Fig. 335 shows 
one way in which the mechanism is actually used. The crank BA is 
in this case the gear M. A is a pin fast to M and carries a block 
which works in a slot in T. M is driven by the pinion K. The link 
H connects N to the ram R which carries the cutting tool. The length 
of the crank BA may be changed by turning the screw S, thus changing 
the length of stroke of the tool. 

Ratio of Time of Cutting Stroke to Time of Return Stroke. Fig. 336 
represents, in diagrammatic form, the mechanism of Fig. 335. The 
driving crank BA turns right-handed as shown by the arrow. When 
the linkage is in the position CNiRi the tool slide R is about to start 
on the cutting stroke. It makes this stroke while CN turns through 
the angle NiCN or BA turns through the angle a. Similarly, R makes 
its return stroke while BA is turning through the angle /5. If BA is 
assumed to turn at a uniform angular speed it follows that 

Time required for cutting stroke _ a 
Time required for return stroke /3 

Angular Speed of Swinging Arm. The angular speed of the swinging 
arm CN for any position as CN 2 may be found by the principle of 
§!228, remembering that the infinite crank at C is always perpendicular 
toCiV. 

From Equation (73) 

Angular speed CN 2 _ Bm 
Angular speed BA CA 2 



268 



ELEMENTS OF MECHANISM 





LINKWORK 



269 



Then if the angular speed of BA 2 is assumed to be one radian per unit 

of time ]$ m 

Angular speed of CN 2 = 77-r- ■ (85) 

Linear Speed of N. Since the linear speed of a point on a swinging 
arm is equal to the angular speed of the arm multiplied by the distance 
of the point from the axis, the linear speed of N2 becomes, from Equa- 
tion (85), B m 

Linear speed iV 2 - CN 2 X 7TT • (86) 



i\ 



! " Intersecting' 
j at 

I 



Xi~ 



h 



<i Cutting Stroke 
. Refurn Stroke 
I 
R1, ! ! <R- 



-"X — X 
\ 

}-V- 




Line of Centers 



Cranio-"" r 



Fig. 336 

Linear Speed of Tool Slide. If the point of intersection of CN 2 and 
the perpendicular to RiR at R 2 (which in this case falls outside the 
limits of the drawing) is represented by the letter 0, then- is the 
instantaneous axis of ^2^2 and 

Linear speed of R 2 _ 0R 2 

Linear speed of N 2 0N 2 



(87) 



270 



ELEMENTS OF MECHANISM 



(88) 



(89) 



By drawing XXi parallel to R2N2, Equation (87) may be written 
Linear speed of R 2 _ KR 2 
Linear speed of N 2 TN 2 

Substituting in (88) the linear speed of N 2 obtained in (86) gives 

t • inn /t»t Bm KR 2 

Linear speed of R 2 = CN 2 X 777- X j^w 

KjA-2 1 1\ 2 

When, as is usually the case, R X R is perpendicular to CB, the triangle 
N 2 Cy, formed by the intersection of R 2 N 2 (produced if necessary), with 
CB produced, is similar to triangle R 2 ON 2 , therefore 

OR 2 Cy 

ON 2 CN 2 
Hence Equation (87) may be written 

Linear speed of R 2 _ Cy 

CN 2 

Bm . . Cy Bm 



Linear speed of N : 
Linear speed of R 2 = CN 2 X 



(90) 



CA 2 X CN 2 



CA, 



xCy. 



(91) 



It should be remembered that Equation (89) is general and that Equa- 
tion (91) applies only when the path of the tool slide is perpendicular 
to the line of centers CB. From each of the Equations (89) and (91) the 
actual linear speed of the tool slide is obtained by multiplying the right- 
hand side of the equation by the angular speed of the driving crank, 
BA, expressed in radians per unit of time. 



Crank 



B "--/e 



#/ 




1 


1 


/ 


C 


1 

i 

1 


Cylinder 


/ Line of Centers 1 




-i 





Fig. 337 

Oscillating Engine. Fig. 337 shows in diagrammatic form a type of 
engine, known as an oscillating engine, which still further illustrates 
the application of the swinging block linkage. Here the crank pin A 



LINKWORK 



271 



is connected directly to the end of the piston rod, and turning is made 
possible by pivoting the cylinder on trunnions at C. The links of the 
equivalent four-bar linkage are indicated in the figure. 

240. Turning Block Linkage. If BA (Fig. 321) is made the fixed 
link, the four-bar linkage becomes that shown in Fig. 338. Here BC is 
a crank turning about B and causing T to turn about A. The ratio of 




\Conn R eot;a 9 



Crank 



V 



Fig. 338 

the angular speed of T to that of A is variable, since the ratio of the 
perpendiculars from B and A to the center line of the theoretical con- 
necting rod varies. 

Whitworth Quick Return is the name given to the linkage when it is 
used as a quick-return mechanism, as in Fig. 339. If the crank BC 
(Fig. 339) turns uniformly right-handed from the position Ci to the 
position C 2 , the slide R will travel from its extreme position at the right 
to the end of its stroke at the left; and while BC turns from C% to Ci, 
the slide R returns 

• Time of cutting stroke of R _ a 
Time of return stroke of R /3 

To locate the center A, given the time ratio of cutting stroke to 
return stroke, the line of centers, the axis B and the crank BC; make 

time of cutting stroke 



the angle CiBA equal to £> where 



time of return stroke 



and 



draw C\A through Ci perpendicular to the line of centers; the point A 



272 



ELEMENTS OF MECHANISM 



is the axis of the link AN. If the stroke of the slide R is not on a line 
passing through A, but below it, as is commonly the case, the time 
ratio of cutting stroke to return stroke would be somewhat different 
from the above. 

The ratio of angular speeds of AN and BC and the linear speed of R 
for any position of the mechanism may be found by a method corre- 




Fig. 339 

sponding to that used for the swinging block quick return, described 
in § 239. 

It will be noticed that this is similar in appearance to the swinging 
block mechanism shown in Figs. 333 and 334, except that in this case 
the driving crank is longer than the line of centers, so that the piece T 




Fig. 340 



makes a complete revolution when BC turns once, whereas the arm T 
in Figs. 333 and 334 merely oscillates. 

Fig. 340 shows in more detail the manner in which this mechanism 
is actually applied. 



t^^a^mmm 



LINKWORK 



273 



241. Sliding-Slot Linkage. If the block, Fig. 321, is fixed to the 
frame so that it can neither turn nor slide, the resulting mechanism is as 
indicated in Fig. 341. The link CB becomes a crank oscillating about 



Gaaaecfi'ng 




Fig. 341 

the fixed axis C. The connecting rod BA may make complete turns 
about the axis A, at the same time that A moves in a straight fine, 
carrying T with it. If BA makes a complete turn relative to A the 
stroke of T is equal to 2 BA. 

Fig. 342 illustrates a manner in which this mechanism may be applied. 
The parts are lettered to correspond with Fig. 341. The worm wheel 



< 





un T 


— - 


1 










1 1 


tfR/S 


[/ 


Or 


.A J A i 


















Fig. 342 



carrying the pin B forms the connecting rod. This wheel may be made 
to rotate about the axis A by a worm keyed to the shaft T. The worm 
and wheel are kept in contact by a piece which supports the bearing of 
A, hangs from the shaft T, and confines the worm between its bearings. 



274 



ELEMENTS OF MECHANISM 



A rotation of the shaft T will turn the worm wheel, causing a recipro- 
cation of the axis A, and consequently of the driving shaft T, through a 
distance equal to twice AB. 

Fig. 342a is another application of this linkage. 




Connect'tnp Rod 



Crank 



Fig. 342a 

242. Expansion of Elements in the Linkages with One Sliding 
Pair. In the preceding discussions no consideration has been given 
to the diameters of the cylindrical pairs of elements included in the 
linkages, nor to the size and form of the links themselves, except as 
they were touched on incidentally in illustrating the manner of apply- 
ing some of the mechanisms, as in Figs. 335, 340 and 342. 

Alterations in the diameters of the pairs do not affect the relative 
motions. Also a change in shape or size of the links does not alter the 
relative motions, so long as the center lines of the elementary links 
remain unchanged, and yet such change may make the action of the 
linkage possible. Since enlargements of the elements of the cylindric 
pairs and changes in the form of the links sometimes conceal the real 
nature of the mechanism and cause much indistinctness, it will be well 
to consider a few cases here. 

The sliding-block linkage of Fig. 332 will first be considered. Each 
of its links is more or less closely connected with its three cylindric 
pairs, and their forms are therefore dependent upon the relative sizes 
of the latter, although this size does not affect the nature of their rela- 
tive motion. Evidently the combination is not altered kinematically, 
if the diameter of the crank shaft is increased so as to include the crank 



LINKWORK 



275 



pin as shown in Fig. 343, where the different links are lettered the same 
as in Fig. 332. S is the enlarged shaft whose center is A. The open 
cylinder of the fixed piece T must be enlarged to the same extent as 
the shaft, so that the pair is still closed. 

This arrangement is used in practice, in some slotting and shearing 
machines, to work a short-stroke pump from the end of an engine 




Fig. 343 

shaft; and in other cases where a short crank forms one piece with its 
own shaft. 

If the crank pin is enlarged until it includes the shaft, the result is 
the common eccentric and eccentric rod shown in Fig. 344. Here 




the eccentric is E and is in reality a circular disk with center at B, 
made fast to the crank shaft so that it revolves with the shaft about 
the axis A. The length of the equivalent crank AB is called the 
eccentricity. This mechanism, which differs in form only from the 
common crank and connecting rod, is much used to operate the valve 
mechanism in steam engines, where it is necessary to obtain a recip- 
rocating motion, often less than the diameter of the engine shaft. The 
part of the rod K which encloses E is called the eccentric stray, and is 
made in two parts, and may be separate from K, the eccentric rod, which is 
usually bolted to one of these parts ; the cylindrical pair is also so shaped 
as to allow no axial motion of K on E. 



276 



ELEMENTS OF MECHANISM 



If the crank pin is still further expanded until it includes the cross- 
head pin, the arrangement shown in Fig. 345 is obtained. In this case, 
the element of the cylindric pair which belongs to the crank AB has 




Fig. 345 

been inverted, and thus made open. The rod BC becomes an eccentric 
disk which swings about the axis of bearing in the piece M, and is always 
in contact with the hollow disk E, carried by the shaft turning in T. 




Fig. 346 



If, instead of enlarging the crank pin to include the crosshead pin, 
the latter is enlarged to include the former) the arrangement shown in 
Fig. 346 is obtained. The rod BC is again an eccentric disk or annular 



LINKWORK 



277 



ring; but it now oscillates in a ring forming part of the piece M, while 
the crank pin drives it by internal contact. In order to make the 
relations of these expansions more clear, fine light lines have been 
drawn in each case, showing the elementary links. The above exhausts 
all the practicable combinations of the three cylindric pairs. 

In Fig. 346 the link K may be replaced by an annular ring containing 
the crank pin and oscillating in a corresponding annular groove in the 
piece M . So long as the center of this ring remains the same as that of 
K, the mechanism remains unchanged, and as the motion of the ring 
is merely oscillatory, only a sector of it need be used and enough of the 
annular groove to admit of sufficient motion of the sector in its swing. 
Fig. 347 represents the arrangement altered in this way, the different 




Fig. 347 

parts being lettered the same as in Fig. 346, BC is still the connecting 

rod replaced by K and its motion as a link in the chain remains the 

same as before, and is completely restrained; the shape of the sector 

always fixes the length of the connecting rod. 

This mechanism is made use of in the Stevenson 

and Gooch reversing gears for locomotives, and 

in other places ; the chains are not there simple, 

but compound. It will be noticed that if the 

radius of the slot is made infinite the result will 

be as in Fig. 325. 

The mechanism shown in Fig. 348, which some- 
times occurs in slotting and metal-punching 
machines, is another illustration of pin expansion. 
The whole forms a sliding-block linkage; the 
link K is formed essentially as in Fig. 347, but 
here the profiles against which it works are 
concave on both sides of the crank pin, the upper 
profile being of large, and the lower of very small, radius, but both forming 
part of the block M . The work is done when the block M is moving 




Fig. 348 



278 ELEMENTS OF MECHANISM 

downwards, and the small radius profile being then in use, the friction 
is reduced. In this case the block M is so enclosed by the guides T 
that the profiles representing the crosshead pin lie entirely within the 
sliding pair, an illustration of how the method of expansion can be 
applied to the fourth or sliding pair. 

The illustrations in Figs. 335, 340 and 342 furnish further examples 
of the expansion or modification of various parts of the linkage in 
order to make the mechanical application possible. 

A change in the shape of an elementary link frequently permits 
motions to take place which are not otherwise possible. In Fig. 349, 




Fig. 349 

for example, a complete rotation of BA to cause a reciprocation of M 
would be possible with the open rod K moving around the fixed shaft 
E, but not with the elementary link BC, shown by a light line. 

243. The Isosceles Linkage. If the link BC (Fig. 321) is made 
equal in length to A B the result is a series of mechanisms analagous to 
those obtained from Fig. 321 by fixing successively the four links. 
The mechanisms obtained, however, have special properties due to the 
equality of AB and BC. Moreover, it will be seen that the four cases 
obtained by fixing the four links are reduced to two since the same 
result, kinematically, is obtained by fixing the block M as by fixing 
the piece T and the axis A ; and by fixing BC as by fixing AB. 

The Isosceles Sliding-Block Linkage. If the piece T is fixed as in 
Fig. 350 the mechanism corresponds to that of Fig. 322. Here (Fig. 
350), if AB is the driver and C starts from the position Ci it will be 
found when the crank AB is at an angle of 90° with ACi (the path 
of C) that C is directly over A and any further rotation of AB will 
cause only a similar rotation of CB. In order to cause C to continue 
in its path from this position it will be necessary to pair points on the 
centrode of BC for this position (when T is fixed) with the correspond- 
ing points on the centrode of T (when BC is fixed) as was done in the 
case of the anti-parallel crank linkage in Figs. 316 and 318. 

The centrode of BC is the circle drawn about A as a center with 
radius 2 AB. This can be seen as follows : 



LINKWORK 



279 



In any position of the linkage, as that occupied in Fig. 350, produce 
A B to meet the perpendicular to AC\, through C, at 0, thus finding the 
instantaneous axis for that position. From B draw Bk perpendicular 

AC 

to ACi) then, since ABC is an isosceles triangle, Ak = kC = -75— Hence, 

AO 

from the similarity of the triangles AOC and ABk, AB = — ~- • This holds 

z 

for every position of the linkage. Therefore the locus of 0, the instan- 
taneous axis of BC, is the circle with radius 2 AB. 

A similar method of reasoning can be followed to show that the cen- 
trode of T, with BC fixed, is a circle about B with radius BA. 



Connecting Rod 




Fig. 350 



From the properties of centrodes previously brought out, if the link 
BC is made fast to a disk of radius A B (= BC) with the point B at its 
center and C on its circumference, and this disk is rolled inside a fixed 
hollow cylinder of twice its own diameter, BC will have the same 
motion that it would if it were the connecting rod of the actual four-bar 
linkage ABC, and C will travel on a diameter of the larger circle. 

It is evident that since BC is a radius of the centrode of T (that is, 
of the disk just referred to) and C has a motion along the diameter 
C1AC2, if BC is prolonged to E, making BE equal to BC, E will travel 
the diameter EaE*, being at A when C is at C h at E s when AB and BC 



280 ELEMENTS OF MECHANISM 

are perpendicular to C2AC1 above C1AC2, at A again when C is at d, 
and at £4 when AB is perpendicularly under C1AC2. 

If now, when the actual linkage is used, with the connecting rod 
prolonged to E, a pin is centered at E and corresponding fixed eyes 
at E z and E 4 as in Fig. 351, a means is provided for causing C to 
continue in its path when AB is in the 90° positions. 




Fig. 351 

It should be noticed that the paths of the points C and E, Fig. 350, 
as shown above, when considered as points of the circumference of the 
smaller circle (that is, on the surface of the centrode of T) are hypo- 
cycloids with the circle AO as the directing circle and circle BO as the 
generating circle. From this it is evident that the prolongation BE 
need not be in the same fine as BC but may be at any angle as at BE 6 , 
provided the eye is properly located. 

If the crank AB turns at uniform angular speed C has harmonic 
motion over the path C1AC2 for C is always found at the foot of the 
perpendicular OC and is always on the line AB produced distant 
2AB from 0. This agrees with the description given for harmonic 
motion in Chapter I. 

In Fig. 352, lettered to correspond to Fig. 350, the actual crank AB is 
omitted and a block is placed at the end of the rod BE to guide E in a 
slot whose center line is a straight line passing through A. The nature 
of the linkage remains the same as in Fig. 350. The imaginary line 
joining B, the middle point of EC, to A, the point of intersection of the 
center lines of the slots is still the theoretical crank and, if motion is 



LINKWORK 



281 



imparted to C, B will move in a circular path about A. The whole 
may be thought of as two four-bar linkages ABCD and ABED h with 
the crank A B common to the two linkages. 

The Elliptic Trammel, which is so commonly used for drawing ellipses, 
is an application of the principle of the isosceles sliding-block linkage. 
Referring to the actual linkage as in Fig. 353, and its equivalent as in 
Fig. 354 (with CBE a straight line), it has been shown that E (Fig. 353) 
moves in a straight line through A and that B (Fig. 354) moves in a 




Line of Centers _W 



~~ >S S9 at infinity 
Fig. 352 



circle about A. If any other point, as P, on the rod CE or CE pro- 
longed, is chosen, P can be shown to move in a path which is an ellipse 
with axes lying along the paths of C and E and with semi-axes equal in 
length to PE and PC. If PE is less than PC the minor axis lies along 
the path of C. If PC is less than PE the minor axis lies along the path 
of E, as in the figure. 

In the elliptic trammel the mechanism is usually applied in the form 
corresponding to Fig. 354 and the ellipse is usually traced by an ad- 
justable point outside of E or C as in the figure; E and C are made so 
that their distance apart is adjustable and they are set one-half the 
difference of the major and minor axes apart. 

An ellipse can be readily drawn by taking a card one corner of which 
shall represent the tracing point P. Points corresponding to the de- 



282 



ELEMENTS OF MECHANISM 



sired positions of E and C are then marked on the edge of the card, 
and by placing these points in successive positions on lines at right 
angles with each other, corresponding to the slots in which the blocks 
in Fig. 354 move, and marking the successive positions of P, will give 
a series of points on the required ellipse. 




Fig. 353 



To prove that the point P moves on an ellipse, let Pn = x; Pr = y; 
PE (semi-major axis) = a; PC (semi-minor axis) = 6. 

The equation of an ellipse referred to the center as the origin is 





x*y 2 - 
a 2 "*" 6 2 


In Figs. 353 and 354 




x Pn A y Pr 
a~PE and b~ PC 


and, since the triangles nPE and rCP are similar, 




Pr nE 




PC PE 


Therefore 


Pn Pr Pn nE 
PE + PC~ PE + PE 


and, in this case, 


~P~n 2 P? Pn^_ nE 2 Pn 2 + n~E 2 
PE 2 PC 2 PE 2 PE 2 PE 2 


Therefore 


x 2 y* 

a? ^ 6 2 ' 



showing that the locus of P is an ellipse. 

By fixing the link BC, Fig. 350, or the equivalent, fixing the centers 
C and E (allowing the blocks to turn), Fig. 352 or Fig. 354, the mechan- 
ism corresponding to the swinging-block linkage, Fig. 332, is obtained. 



LINKWORK 



283 



Two examples will be considered in which this linkage is expanded in 
the manner suggested in Figs. 352 and 354. 

The Elliptic Chuck depends upon the principle proved for the elliptic 
trammel and upon the principle, previously referred to, that the relative 
motions of the parts of a linkage are independent of the fixedness of 
the links. 

Now in drawing an ellipse with a trammel, the paper is fixed, and 
the pencil is moved over it; but in turning an ellipse in a lathe, the 
tool, which has the same position as the pencil, is fixed, and the piece to 
be turned should have such a motion as would compel the tool to cut 




This is accomplished in the elliptic chuck, in which the 
spindle of the lathe, with a block on it, corresponds to the axis E of 
Fig. 354. In another fixed bearing whose axis corresponds to C is 
another shaft having a block on it. The point of the cutting tool is in 
a fixed position corresponding to P. 

The piece carrying the work corresponds to T and has two slots 
at right angles, sliding over the blocks on the spindle and the axis C. 
The turning of the spindle causes the point of intersection of the center 
fines of these slots to move in a circle about the axis of the spindle 
and the whole piece to have such a motion that the point of the tool 
cuts an ellipse from the material attached to it. 



284 



ELEMENTS OF MECHANISM 



The Oldhams Coupling shown in Fig. 355 is an interesting example of 
this form of linkage. The axis E of the upper shaft corresponds to E 
in Fig. 354 and the slotted disk on this shaft corresponds to the block 
at E; similarly, with the shaft at C. The intermediate disk T with 
two projections at right angles across its diameters replaces the cross T. 




Fig. 355 

The object of the device is to connect two parallel shafts placed a 
short distance apart so as to communicate uniform rotation from one 
to the other. 

If the link AB, Fig. 350, is made the stationary link the result is the 
same, kinematically, as the linkage discussed in the last two illustra- 




Fig. 356 



tions. The details of application are somewhat different, 
shows the mechanism. 



Fig. 356 



LINKWORK 



285 




Fig. 357 



If T is on a shaft centered at A the crank BC, through the arm T, 
will cause the shaft at A to turn at an angular speed equal to one-half 
its own. There may be two arms BC and BE with corresponding 
slots, the result being the equivalent of a two-toothed wheel driving, 
internally, one of four teeth. If there were three arms and three slots 
the equivalent gears would have three and six teeth. 

244. Other Forms of Linkwork with Two Sliding Pairs. Fig. 357 
shows a combination of an eccentric circular 
disk A, and a sliding piece C, moving through 
fixed guides, one of which is shown at D. A 
uniform rotation of A about the axis a will 
give harmonic motion to C. This can be 
shown by noticing that the distance which C 
has moved from its highest position is 



cd = ef = ab (1 — cos 0), 

which is the equation for simple harmonic 

motion where 2 ab is the stroke of the slide. 
This mechanism can also be found by an 

expansion of the crank pin B, Fig. 325, until it 

includes the shaft A, the slot being correspondingly enlarged, ana then 

after turning the figure through 90°, omitting the lower part of the 

cross, allowing the pairing to be by force-closure. 

The Swash-Plate. The apparatus shown in Fig. 358, known as a 

swash-plate, consists of an elliptical plate A set obliquely upon the 
shaft S, which by its rotation causes a sliding bar C to 
move up and down, in a line parallel to the axis of the 
shaft, in the guides D, the friction between the end of 
the bar and the plate being lessened by a small roller 0. 
When a roller is used, the motion of the bar C is approxi- 
mately harmonic — the smaller the roller the closer the 
approximation. If a point is used in place of the 
roller, the motion is harmonic, which can be shown as 
follows : 

Since the bar C remains always parallel to the axis of 
the shaft, the path of the point 0, projected upon an 

imaginary plane through the lowest position of and perpendicular 

to the shaft S, will be a circle, and the actual path of on the plate 

A will be an ellipse. 

In Fig. 359 let eba represent the angular inclination of the plate to 

the axis of the shaft, ab the axis of the shaft, eof the actual p'ath of the 

point o on the plate, and the dotted circle erd the projection of this 




Fig. 358 



286 



ELEMENTS OF MECHANISM 



path upon a plane through e (the lowest position of o) perpendicular 
to the axis ab. 

Draw om perpendicular to ef, or perpendicular to the plane erd, and 
rn perpendicular to ed, the diameter of the circle erd. Join mn, and 
suppose the plate to rotate through an angle ear = 6, and thus to carry 
the point o through a vertical distance equal to or. 

Then 



or = mn = ab X 




Fig. 359 



/ mn _ en\ 
\ ab eaj 

= ab{l — cos0), 



or the same formula as was derived in the case of harmonic motion. In 
this case ab represents the length of the equivalent crank, and is equal 
in length to one-half of the stroke of the rod C. 

245. The Conic Four-Bar Linkage. If the axes of the four cylin- 
dric pairs of the four-bar linkage are not parallel, but have a common 
point of intersection at a finite distance, the chain remains movable 
and also closed (Fig. 360). The lengths of the different links will now 
be measured on the surface of a sphere 
whose center is at the point of inter- 
section of the axes. The axoids will no 
longer be cylinders, but cones, as all the 
instantaneous axes must pass through 
the common point of intersection of the / 

pin axes. / ^ 

The different forms of the cylindric J£zZl 
linkage repeat themselves in the conical 
one, but with certain differences in their 
relations. The principal difference is in the relative lengths of the links, 
which would vary if they were measured upon spherical surfaces of 
different radii, the links being necessarily located at different distances 
from the center of the sphere in order that they may pass each other in 
their motions. The ratio, however, between the length of a link and its 
radius remains constant for all values of the radius, and these ratios are 
merely the values of the circular measures of the angles subtended by 
the links. In place of the link lengths, the relative magnitudes of these 
angles can be considered. 




LINKWORK 



287 



The alterations in the lengths of the links will now be represented by 
corresponding angular changes. The infinitely long link corresponds 
to an angle of 90°, as this gives motion on a great circle which corre- 
sponds to straight-line motion in the cylindric linkages. 

Fig. 361 shows plan and elevation of a conic four-bar linkage ABCD, 
the link AB turning about A, and, for a complete turn, causing an 
oscillation of the link CD about D through the angle 0, shown in the 
elevation. In the figure each of the links BC and CD subtends 90°, 
while the link AB subtends about 30°. Varying the angles which the 
links subtend will, of course, vary the relative 
motions of AB and CD. 

246. Hooke's Joint. If in Fig. 361 each 
of the links AB, BC, and CD is made to sub- 
tend an angle of 90°, AB and CD will each 
make complete rotations. This mechanism, 
known as a Hooke's joint, is represented by 
Fig. 362, A and D are the two intersecting 
shafts, and the links AB and CD, fast to the 
shafts A and D re- 
spectively, subtend 
90°, while the con- 
necting link BC also 
subtends 90°. 

In order to make 
the apparatus 
stronger and stiff er, 
two sets of links are 
used, and the link 
CB is continued 
around as shown, 
thus giving an 
annular ring joining the ends of the double links CDC and BAB'. This 
ring is sometimes replaced by a sphere into which the pins C, B, C , 
and B' are fitted, or by a rectangular cross with arms of a circular 
section working in the circular holes at B, C, C , and B' . Or, the arms 
BAB X and CACi may be paired with grooves cut in a sphere in planes 
passing through the center of the sphere and at right angles to each 
other. Such forms of Hooke's joint are much used. 

Relative Motion of the Two Connected Shafts. — Given the angular 
motion of AB, to find the angle through which CD turns. Fig. 363 
shows a plan and elevation of a Hooke's joint, so drawn that the axis 
A is perpendicular to the plane of elevation. If the link AB is turned 





Fig. 361 



Fig. 362 



288 



ELEMENTS OF MECHANISM 



through an angle 6, it will be projected in the position AB\. The path 
of the point C will be on a great circle in a plane perpendicular to the 
axis D, which will appear in the elevation as the ellipse BCE. The 
point C will then move to d, found by making 
the angle BiACi equal to 90°, for the link BC 
subtends 90°, and since the radius from B to 
the center of the sphere is always parallel to 
the plane of elevation, its projection and that 
of the radius from C will always be at right 
angles. The projected position of the linkage 
after turning A through the angle 6 will be 
AB1C1D. To find the true angle through 
which the link CD and the shaft D have 
turned, swing the ellipse BCE with the axis 
D, until D is perpendicular to the plane of 
elevation, when the points C and Ci will be 
found at C and C\ , respectively, giving the 
angle C\AC = </> as the true angle through 
which the axis D has turned. Or the arm 
DCi may be revolved until shaft D is perpen- 
dicular to the horizontal plane, giving CiBCi 
= (j>, as shown in the plan. 

It is evident from the above that two 
intersecting shafts connected by a single 
Hooke's joint cannot have uniform motions. 
If, however, two joints are used to connect 
two parallel or intersecting shafts, they may be so arranged that they 
will have uniform motions. 

247. Double Hooke's Joint. Two parallel or intersecting shafts may 
be connected by a double Hooke's joint and have uniform motions, provided 
that the intermediate shaft makes equal angles with the connected shafts, 
and that the links on the intermediate shaft are in the same -plane. Fig. 
364 gives a plan and elevation of two shafts so connected, and the posi- 
tion after turning through an angle 6. It is evident that one joint just 
neutralizes the effect of the other. 

The term universal joint is often used to designate the above- 
described mechanism. 

248. Angular Speed Ratio in a Single Hooke's Joint. Fig. 365 
reproduces the elevation given in Fig. 363, which shows the angles 
6 and <£ through which A and D move respectively. The angle EAC 
= a will be the true angle between the planes in which the paths of the 
points B and C lie; to find the angle analytically in terms of 6 and a, 
we have, from Fig. 365, 




LINKWORK 



289 



, Ci'G CiF . a CiF 
tan * = liG = AG W= lF 



tan^CiF AF = AF AC _ AC _ 
tan0 AG'dF AG AC AE C0S a; 

.'. tan 6 = tan cos a. 



(92) 



To obtain the velocity ratio, differentiate equation (92), remembering 
that cos a is a constant ; then 



d4 _ sec 2 6 
dd sec 2 4> 



1 + tan 2 6 

a = r~n — n cos a - 

1 + tan- 4> 



(93) 




Fig. 364 



If we eliminate <f> and from equation (93), by use of equation (92) 
we shall obtain 

d4> cos a 



d8 1 — sin 2 sin 2 a 

_ 1 — cos 2 <j) sin 2 a 
cos a 



(94) 
(95) 



290 



ELEMENTS OF MECHANISM 



Assume AB and CD the starting positions of the arms AB and CD 
respectively; then equations (94) and (95) will have minimum values 
when sin 6 = and cos </> = 1 ; this will happen when 6 and <£ are 0° and 

180°, giving ^- = cos a in both cases. Thus the minimum velocity ratio 
dv 

occurs when the driving arm is at AB and AB 2 , the corresponding posi- 
tions of the following arm being CD and 
CJ). Maximum values occur when sin 

= 1 and cos = 0; then ■$- = > which 

ad cos a 

will happen when 6 and cf> are 90° and 270°, 

the corresponding positions of the driving 

arm being AB S and AB4. 

Hence in one rotation of the driving 

shaft the velocity ratio varies twice be- 

1 




tween the limits — t — and 
cos a 



cos a; and be- 



Fig. 365 



tween these points there are four positions 

where the value is unity. 

If the angle a increases, the variation in the angular velocity ratio of 

the two connected shafts also increases; and when this variation becomes 

too great to be admissible in any case, other arrangements must be 

employed. 



CHAPTER XII 
STRAIGHT-LINE MECHANISMS — PARALLEL MOTIONS 

249. A Straight-line Mechanism is a linkage designed to guide a 
reciprocating piece either exactly or approximately in a straight line, 
in order to avoid the friction arising from the use of straight guides. 
Some straight-line mechanisms are exact, that is, they guide the recipro- 
cating piece in an exact straight line; others, which occur more fre- 
quently, are approximate, and are usually designed so that the middle 
and two extreme positions of the guided point shall be in one straight line, 
while at the same time care is taken that the intermediate positions 
deviate as little as possible from that line. 

250. Peaucellier's Straight-line Mechanism. Fig. 366 shows a 
linkage, invented by M. Peaucellier, for describing an exact straight 
line within the limits of its motion. 

It consists of eight links joined at their ends. Four of these links, 
A, B, C, and D, are equal to each other and form a cell; the two equal 



\ < 




Fig. 366 



links E and F connect the opposite points of the cell, a and e, with the 
fixed center of motion d; the link G = %bd oscillates on the fixed center 
c, cd thus forming the fixed link equal in length to G. 

If now the linkage be moved within the working limits of its con- 
struction (that is, until the links B and G, and C and G come into line 
on opposite sides of the center line of motion cd) , the cell will open and 

291 



292 ELEMENTS OF MECHANISM 

close; the points a and e will describe circular arcs about d, and b about c. 
Finally, the point p will describe a straight line ss perpendicular to the 
line of centers cd. 

To prove this, move the linkage into some other position, as piaib x cd. 
(It is to be noticed that since the links A and D, B and C, and E and F 
always form isosceles triangles with a common base, a straight line from 
p to d will always pass through b.) If the line traced by the point p is 
a straight line, the angle pipd will be 90°. The angle bbid is 90°, since 
be = cd = bic; therefore the triangles pipd and bbid would be similar 
right triangles, and we should have 

pd _ bid 
pid bd 

To prove that ss is a straight line it is necessary to show that the 
above relation exists in the different positions of the linkage. In Fig. 366 

E 2 = aJ 2 +(bf + bd)^ 
B* = aj 2 + Vf; 
.'. E*-B 2 = 2 (bf) (bd) + bd 2 = bd (bd + 26/). 

But, since the links A and B are equal, the triangle pab is isosceles and 
the base pb = 2 bf. 

.'. E 2 - B 2 = (bd) (pd). (96) 

By the same process, when the linkage is in any other position, as 
pia-fcxcd, we should have 

E£ - 5i 2 = (bid) (pid). (97) 

Equating equations (44) and (45), 

(bd) (pd) = (hd) (pd), 
pd bid 
pid bd 

which proves that the path of the point p is on the straight line ss. 

If the relation between the links cd and be be taken different from 
that shown (Fig. 368), the points b and p, sometimes called the poles 
of the cell, will be found to describe circular arcs whose centers are on 
the line passing through c and d; in the case shown, one of these cir- 
cular arcs has a radius infinity. 

251. Scott Russell's Straight-line Mechanism. This mechanism, 
suggested by Mr. Scott Russell, is an application of the isosceles sliding- 
block linkage. 

It is made up of the links db and pc, Fig. 367. The link ab, centered 
at a, is joined to the middle point b of the link pc, and ab, be and pb 
are taken equal to each other; and the point c is constrained to move 



STRAIGHT-LINE MECHANISMS 



293 



in the straight line ac by means of the sliding block. In this case the 
motion of the sliding block c is slight, as the entire motion of p is seldom 
taken as great as cp. 

To show that the point p describes a straight line ppip* perpendicular 
to ac through a, a semicircle may be drawn through p and c with b as a 
center; it will also pass through a so that pac will be a right angle- 
therefore the point p is on ap, 
which is true for all positions 
of p. 

The point a should be located 
in the middle of the path or 
stroke of p. The motion of c 
may then be found by the 
equation 



cci = cp — v cp 



ap 




Fig. 367 



where ap is the half -stroke of p. 

Approximate straight-1 i n e 
motions somewhat resembling 
the preceding may be obtained by guiding the link cp entirely by 
oscillating links, instead of by a link and slide. 

1° In the link cp (Fig. 367) choose a convenient point e whose mean 
position is e 1} and whose extreme positions are e and e 2 . Through these 
three points pass a circular arc, ee-ify, the center of which / will be found 
on the line ac. Join e and / by a link ef, and the two links ab and ef 
will so guide pe that the mean and extreme positions of p will be found 
on the line pp 2 , provided suitable pairs are supplied to cause passage by 
the central position. 

2° The point c may be made to move very nearly in a straight line 
cc\ by means of a link cd centered on a perpendicular erected at the 
middle point of the path of c. The longer this link the nearer the path 
of c will approach a straight line. 

This straight-line motion has been applied in a form of small sta- 
tionary engines, commonly known as grasshopper engines, where cbp 
(Fig. 367), extended beyond p, forms the beam of the engine,, its right- 
hand end being supported by the link cd. The piston-rod is attached, 
by means of a crosshead, to the point p, which describes a straight line, 
and the connecting-rod is attached to a point in the line cp produced, 
both piston-rod and connecting-rod passing downward from cp. In 
this case it will be noticed that the pressure on the fulcrum c, of the 
beam, is equal to the difference of the pressures on the crosshead pin 
and crank-pin instead of the sum, as in the ordinary form of beam engine. 



294 



ELEMENTS OF MECHANISM 



In this second form of motion it is not always convenient to place the 
point a in the line of motion pp 2 , and it is often located on one side, as 
shown in Fig. 368 

P 




Fig. 368 



The proportions of the different links which will cause the point p 
to be nearly on the straight line at the extreme positions and at the 
middle may be found as follows: 

Let pg be one-half the stroke of the point p, and let the angle bac = 0, 
and bca = pbe = 4>. In this extreme position we may write 
ag = af - fg = af - be 
= ab cos 9 — pb cos 



ab 



(l - 2 sin 2 Q - pb(l - 2sin 2 |Y 



2/ ' \ 2, 

But if the links are taken long enough, so that for a given stroke the 
angles 9 and <f> are small, then sin 9 = 9, nearly, and sin <f> = <j>, nearly, 
and 



(98) 





ag = aMl--j- 


4-1) 




= ab — pb — a& p + pb-~- 


If the linkage is 


now placed in its mid-position, 




ag = ab — 


pb. 


Equating Equations (98) and (99), 






ab^ = pb- 


> 


or 

But in the triangle 


ab 2 
pb ~ 9 2 ' 
abc 

ab sin $ 
be sin 9 9 


nearly; 




ab ab 2 
Pi> be 


(ab) (pb) = I 



bc\ 



(99) 



(100) 



(101) 




Fig. 369 



STRAIGHT-LINE MECHANISMS 295 

Hence the links must be so proportioned that be is a mean propor- 
tional between ab and pb, which also holds true when the path of p 
falls to the left of a instead of between a and c. 

As an example of the case where the path of the guided point falls 
to the left of a we have the straight-line motion of the Thompson 
steam-engine indicator, Fig. 374. 

252. Watt's Straight-line Mechanism. Fig. 369 shows a Watt 
straight-line mechanism. Here the two links ad and be connected by 
the link ab oscillate on the fixed 
centers d and c, and any point, as 
p, in the connecting link ab will 
describe a complex curve. If the 
point p be properly chosen, a 
double-looped curve will be ob- 
tained, two parts of which are 
nearly straight lines. In designing 
such a motion it is customary to 
use only a portion ef of one of the 
approximate straight lines, and to 
so proportion the different links 
that the extreme and middle points e, f, and p shall be on a line per- 
pendicular to the center lines of the levers ad and be in their middle 
positions, when they should be taken parallel to each other. 

The linkage is shown in its mid-position by dabc, Fig. 370, and in the 
upper extreme position by dajjic, where ppi is to be one-half the stroke 
of p. Given the positions of the links ad and be when in their mid- 
position, the axes c and d, the line of stroke ss, and the length of the 
stroke desired; to find the points a and 6, giving the link ab, and to 
prove that the point p, where ab crosses ss, will -be found on the line ss 
when it is moved up (or down) one-half the given stroke. Lay off 
on ss from the points- g and h, where the links ad and be cross the fine 
ss, one-quarter of the stroke, giving the points k and I; connect these 
points with the axes d and c respectively; draw the lines akai and blb\ 
perpendicular to dk and el respectively, making aa\ = 2ak and bbi = 2 bl; 
then if the link centered at d were ad, it could swing to aid, and similarly 
be could swing to bic. By construction kg = | stroke, and aai = 2 ak; 
therefore a x e = \ stroke. Similarly b\f = \ stroke, which would make 
the figure eaibif a parallelogram, and aib x would equal ef. But ef is 
equal to ab, since bh = hf and ag = ge. Therefore, if the linkage is 
dbac, it can occupy the position daibic; and since ap = ep = aipi, and 
ppi = eai = | the stroke, the point d will be at p x and \ the stroke 
above p. 



296 



ELEMENTS OF MECHANISM 



To calculate the lengths of the links, given dg, ch, and gh, and the 
length of stroke S. Since the chord aa x of the arc through which a 
would move is bisected at right angles by the line dk, 



Similarly, 





gk 2 = (ag) (dg) 


16' 


ag -- 


S 2 
= zttt-t and ad 
16 dg 


= dg 


be-- 


S 2 
16 ch 




ab = 


= (gh 2 + (ag + bh) 2 f 




^ 2+ (i^ + 


S 2 , 
16^ 



+ 



IQdg 




Fig. 370 



To determine the position of the point p we have from the figure 

J ap :bp = ag : bh. (102) 



S 2 



S 2 



from which 



ap : ab = ch : ch + 
bp : ab = dg : ch + 



ch : dg, 



STRAIGHT-LINE MECHANISMS 



297 



from which the position of the guided point p can be calculated, 
is very often the case, ad = be, then 

£ 2 



If, as 



ad = be = 



lQdg 



and bp : ab = dg : 2 dg, 

or bp = \ ab, 

and the point p is thus at the middle of the link ab. 

This mechanism may be arranged as shown in Fig. 371, where the 
centers c and d are on the same side of the line of motion. The graphical 

9 h 




Fig. 371 

solution is the same as in Fig. 370, with the result that p is found where 
ab extended crosses the line of stroke ss, and, as before, it can be shown 
that if p is moved up one-half the given stroke, it will be found on the 
line of stroke ss. 

In Fig. 370, letting the angle adai = 6 and bcbi = <p, we have, from 
Equation (102), 

ap _ ag _ ae _ ad (1 — cos 6) 2 

bp bh bf ~ be (1 — cos 0) _ r~ 2 • ,</> ' 

be sm 2 ^ 



298 ELEMENTS OF MECHANISM 

which may be written 



ad 2 sin 2 - 
2 

bp ad r"2 • «, d> 
be sm 2 - 



ap _ be ^ 2 

x — 



2 

But ad sin = be sin 0; and since the angles 8 or <j> would rarely exceed 
20°, we may assume that 

ad sin - = be sin -• 

••• i -£'""*. < 103 > 

or the segments of the link are inversely proportional to the lengths of 
the nearer levers, which is the rule usually employed when the extreme 
positions can vary a very little from the straight line. When the levers 
are equal this rule is exact. 

253. The Pantograph. The pantograph is a four-bar linkage so 
arranged as to form a parallelogram abed, Fig. 372. Fixing some point 
in the linkage, as e, certain other points, as/, g, and h, will move parallel 
and similar to each other over any path either straight or curved. These 
points, as f, g, and h, must lie on the same straight line passing through 
f 




Fig. 372 

the fixed point e, and their motions will then be proportional to their distances 
from the fixed point. To prove that this is so, move the point / to any 
other position, as fi, the linkage will then be found to occupy the posi- 
tion aihcidi. Connect /i with e; then h h where fie crosses the link 61C1, 
can be proved to be the same distance from c\ that h is from c, and the 
line Mi will be parallel to ffi. 



STRAIGHT-LINE MECHANISMS 299 

In the original position, since fd is parallel to he, we may write 

fd de /e_ 
he ce he 

In the second position, since fid x is parallel to hid and since fie is drawn 
a straight line, we have 

fidi = di0 = M m 

hid de he 

Now in these equations — = —', therefore -r- = ^-^', but fd = Adi, 
H ce cie he hid J J ' 

which gives he — hid, which proves that the point h has moved to hi. 
Also j- = jr-> from which it follows that ff x is parallel to hh h and 

ffi = fe = de 
hhi he ce 

or the motions are proportional to the distances of the points / and h 
. from e. 

To connect two points, as a and b, Fig. 373, by a pantograph, so that 
their motions shall be parallel and similar and in a given ratio, we have, 
first, that the fixed point c must be on the straight line ab continued, 




Fig. 373 

and so located that ac is to be as the desired ratio of the motion of a to b. 
After locating c, an infinite number of pantographs might be drawn. 
Care must be taken that the links are so proportioned as to allow the 
desired magnitude and direction of motion. 

It is interesting to note that if b were the fixed point, a and c would 
move in opposite directions. It can be shown as before that their 
motions would be parallel and as ab is to be. 

The pantograph is often used to reduce or enlarge drawings, for it 
is evident that similar curves may be traced as well as straight lines. 
Also pantographs are used to increase or reduce motion in some definite 
proportion, as in the indicator rig on an engine where the motion of 
the crosshead is reduced proportionally to the desired length of the 
indicator diagram. When the points, as/ and h (Fig. 372), are required 
to move in parallel straight lines it is not always necessary to employ 
a complete parallelogram, provided the mechanism is such that the 



300 



ELEMENTS OF MECHANISM 



points / and h are properly guided. Such a case is shown in Fig. 374, 
which is a diagram of the mechanism for moving the pencil on a Thomp- 
son steam-engine indicator. The pencil at /, which traces the diagram 
on a paper carried by an oscillating drum, is guided by a Scott-Russell 
straight-line motion abed so that it moves nearly in a straight line ss 
parallel to the axis of the drum, and to the center line of the cylinder 
tt. It must also be arranged that the motion of the pencil / always 
bears the same relation to the motion of the piston of the indicator on 
the line tt. To secure this draw a line from f to d and note the point e 
where it crosses the line tt. Then e will be a point on the piston-rod, which 
rod is guided in an exact straight line by the cylinder. If now the link 
eh is added so that its center line is parallel to cd, we should have, 




Fig. 374 



assuming / to move on an exact straight line, the motion of / parallel to 
the motion of e and in a constant ratio as cf : eh or as df : de. This can 
be seen by supposing the link eg to be added, which completes the 
pantograph dgehef. If eg were added, the link ab could not be used, as 
the linkage abedf does not give an exact straight-line motion to/. For 
constructive reasons the link eg is omitted; a ball joint is located at e 
which moves in an exact straight line, and the point / is guided by the 
Scott-Russell motion, the error in the motion being very slight indeed. 
Slides are often substituted, in the manner just explained, for links 
of a pantograph, and exact reductions are thereby obtained. In Fig. 
375 the points / and h are made to move on the parallel lines mm and nn 



STRAIGHT-LINE MECHANISMS 



301 



respectively. Suppose it is desired to have the point h move § as 
much as /. Draw the line fhe and lay off the point e so that eh : ef 
= 1:3; draw a line, as ed, and locate a point d upon it which when 
connected to / with a link df will move 
nearly an equal distance to the right 
and left of the line ef and above and 
below the line mm for the known motion 
of /. Draw ch through h and parallel 
to df. The linkage echdf will accomplish 
the result required. The dotted link ah 
may be added to complete the panto- 
graph, and the slide h may then be 
removed or not as desired. The figure 
also shows how a point g may be made 
to move in the opposite direction to / 
in the same ratio as h but on the line 
Wifti, the equivalent pantograph being drawn dotted. The link ed is 
shown in its extreme position to the left by heavy lines and to the right 
by light lines. 

254. Combination of Watt's Straight-line Mechanism with a Pan- 
tograph. Watt's straight-line mechanism has been much used in 
beam engines, and it is generally necessary to arrange so that more 
than one point can be guided, which is accomplished by a pantograph 
attachment. 

In Fig. 376 a straight-line mechanism is arranged to guide three 
points p, pi, and p 2 in parallel straight lines. The case chosen is that 





of a compound condensing beam engine, where P% is the piston rod of 
the low-pressure cylinder, Pi that of the high-pressure cylinder, and P 



302 ELEMENTS OF MECHANISM 

the pump rod, all of which should move in parallel straight lines, per- 
pendicular to the center line of the beam in its middle position. 

The fundamental linkage dabc is arranged to guide the point p as 
required; then adding the parallelograms astb and ap 2 rb, placing the 
links st and p 2 r so that they pass through the points p\ and p 2 respect- 
ively, found by drawing the straight line cp and noting points pi and 
p 2 where it intersects lines Pi and P 2 , we obtain the complete linkage. 
The links are arranged in two sets, and the rods are carried between 
them; the links da are also placed outside of the links p 2 a. When the 
point p falls within the beam a double pump-rod must be used. The 
linkage is shown in its extreme upper position to render its construction 
clearer. 

The various links are usually designated as follows: cr the main 
beam, ad the radius bar or bridle, p 2 r the main link, ab the back link, 
and p 2 a the parallel bar, connecting the main and back links. 

In order to proportion the linkage so that the point p 2 shall fall at 
the end of the link rp 2 we have, by similar triangles cbp and crp 2 , 

cb :bp = cr : rp 2 = cr : ab. 

ab X cb 
.'. cr = — t 

bp 

The relative stroke S of the point p 2 and s of the point p are expressed 
by the equation 

S : s = cp 2 : cp = cr : cb. 

If we denote by M and N the lengths of the perpendiculars dropped 
from c to the lines of motion P 2 and P respectively, then 

S :s = M :N 
M N 

and S = s ff'> S = S W' (104) 

The problem will generally be, given the centers of the main beam 
c and bridle d, the stroke S of the point p 2 , and the paths of the guided 
points p, p\, and p 2 , to find the remaining parts. The strokes of the 
guided points can be found from Equation (104) and then the method of 
§ 252, Fig. 370, can be applied. 

255. Robert's Approximate Straight-line Mechanism. This might 
also be called the W straight-line mechanism, and is shown in Fig. 377. 
It consists of a rigid triangular frame abp forming an isosceles triangle 
on ab, the points a and b being guided by links ad = be = bp, oscillat- 
ing on the centers d and c respectively, which are on the line of motion dc. 

To lay out the motion, let dc be the straight line of the stroke along 
which the guided point p is to move approximately, and p be the mid- 



STRAIGHT-LINE MECHANISM 



303 



is the rigid triangular 




Fig. 377 



die point of that line. Draw two equal isosceles triangles, dap and cbp; 

join ab, which must equal dp = pc. Then a 

frame, p the guided point, and d and c are 

the centers of the two links. The extreme 

positions when p is at d and c are shown at 

da±a 2 and ca 2 b 2 , the point a 2 being common to 

both. The length of each side of the triangle, 

as ap = da, should not be less than 1.186 dp, 

since in this case the points ca 2 ai and da 2 b 2 lie 

in straight lines. It may be made as much 

greater as the space will permit, and the greater it is the more accurate 

will the motion be. The intermediate positions between dp and cp vary 

somewhat from the line dc. 

256. Tchebicheff's Approximate Straight-line Mechanism. Fig. 
378 shows another mechanism giving a close approximation to a straight- 
line motion invented by Prof. Tchebicheff. 

The links are made in the following proportion: If cd = 4, then 
ac = bd = 5 and ab = 2. The guided point p is located midway be- 
tween a and b on t he link a b and is distant from cd an 
amount equal to V5 2 — 3 2 = 4. When the point p 
moves to p h directly over d, dpi = dbi — b x p x = 5 — 1 
= 4. Thus the middle and extreme positions of p, as 
shown, are in line, but the intermediate positions will 
be found to deviate slightly from the straight line. 
To render the range of motion shown on the figure 
possible the links ac and bd would need to be offset. 

257. Parallel Motion by Means of Four-bar Linkage. The parallel 
crank mechanism, § 234, Fig. 313, is very often used to produce parallel 
motions. The common parallel ruler, consisting of two parallel straight- 
edges connected by two equal and parallel links is a familiar example 
of such application. A double parallel crank mechanism is applied in 
the Universal Drafting-machine, extensively used in place of T square 
and triangles. Its essential features are shown in Fig. 379. The clamp 
C is made fast to the upper left-hand edge of the drawing-board and 
supports the first linkage abed. The ring cedf carries the second linkage 
efhg, guiding the head P. The two combined scales and straight- 
edges A and B, fixed at right angles to each other, are arranged to 
swivel on P, and by means of a graduated circle and clamp-nut may 
be set at any desired angle, the device thus serving as a protractor. 
The fine lines show how the linkages appear when the head is moved 
to Pi, and it is easily seen that the straight-edges will always be guided 
into parallel positions. 




Fig. 378 



304 



ELEMENTS OF MECHANISM 




Fig. 379 





Fig. 380 



Fig. 381 



STRAIGHT-LINE MECHANISMS 



305 



258. Parallel Motion by Cords. Cords, wire ropes, or small steel 
wires are frequently used to compel the motion of long narrow car- 
riages or sliders into parallel positions. In Fig. 380 the slider R has at 
either end the double-grooved wheels E and F. A cord attached to the 
hook A passes vertically downward under F, across over E, and down- 
ward to the hook C. A similarly arranged cord starts from B, passes 
around E and F, using the remaining grooves, and is made fast to 
hook D. On moving the slider downward it will be seen that for a 
motion of 1" the wheel F will give out 1" of the rope from A and take 
up 1" of rope from D, which is only possible when E takes up 1" from 
C and gives out 1" to B. Thus the 
slider R is constrained to move into 
parallel positions. In practice turn- 
buckles or other means are provided to 
keep the cords taut. 

Figs. 381 and 382 show two other 
arrangements which will accomplish the 
same purpose. In Fig. 381, sometimes 
applied to guide straight-edges on 
drawing-boards, the cords or wires cross 
on the back side of the board where 
the four guide-wheels are located and 
the straight-edge R is guided by special 
fastenings E and F, passing around the 
edges of and under the board. By 
making one of these fastenings movable the straight-edge may be ad- 
justed. Fig. 382 shows a similar device that might be applied on a 
drawing-board. Here the wires are on the front of the board and are 
arranged to pass under the straight-edge in a suitable groove. The 
turnbuckle T serves to keep the wires taut, and the slotted link S allows 
adjustment. 

The device shown in Fig. 380 is often known as a squaring band and 
is applied in spinning-mules' and in some forms of travelling cranes. 





■—-a r 












02 R IC 






Of eCj 






do co 





Fig. 382 



CHAPTER XIII 

MISCELLANEOUS MECHANISMS — AGGREGATE COMBINA- 
TIONS — PULLEY BLOCKS — INTERMITTENT MOTION 



259. Aggregate Combinations is a term applied to assemblages of 
pieces in mechanism in which the motion of the follower is the result- 
ant of the motions given to it by more than one driver. The number of 
independently-acting drivers which give motion to the follower is gen- 
erally two, and cannot be greater than three, as each driver determines 
the motion of at least one point of the follower, and the motion of three 
points in a body fixes its motion. 

By means of aggregate combinations we may produce very rapid or 
slow movements and complex paths, which could not well be obtained 
from a single driver. 

The epicyclic gear trains discussed in Chapter VII in reality come 
under the heading of aggregate combinations. 

260. Aggregate Motion by Linkwork. Figs. 383 and 384 repre- 
sent the usual arrangement of such a combination. A rigid bar ab has 

two points, as a and b, each con- 
nected with one driver, while c 
may be connected with a follower. 
Let aai represent the linear velocity 
of a, and 66i the linear velocity of 
b: to find the linear velocity of c. 
Consider the motions to take place 
separately; then if b were fixed, 
the linear velocity aa x given to a 
would cause c to have a velocity 
represented by cc x . Considering a 
as fixed, the linear velocity bbi at b 
would give to c a velocity cc 2 . The 
aggregate of these two would be 
the algebraic sum of cci and cc%. 

In Fig. 383 we have cci acting to the left, while cc 2 acts to the right; 

therefore the resulting linear velocity of c will be pes = cc\ — cc 2 acting 

306 



M. 



\\ 



— f-Oc 




b$s— °b 

\ 



10 



Fig. 383 



Fig. 384 



MISCELLANEOUS MECHANISMS 



307 



to the left, since cci > cci. In Fig. 384, where both cci and CC2 act to 
the left, the result is cc 3 = cc\ + cc 2 acting to the left. It will be seen 
that the same results could have been obtained by finding the instan- 
taneous axis o of ab in each case, when we should have linear velocity 
c : linear velocity a = co : ao. 

In many cases the lines of motion are not exactly perpendicular 
to the link, nor parallel to each other, neither do the points a, b, and c 
necessarily he in the same straight line, but often the conditions are 
approximately as assumed in Figs. 383 and 384, so that the error intro- 
duced by so considering them may be sufficiently small to be practically 
disregarded. 

As examples of aggregate motion by linkwork we have the different 
.forms of link motions as used in the valve gears of reversing steam 
engines. Here the ends of the links are driven by eccentrics, and the 
motion for the valve is taken from some intermediate point on the link 
whose distance from the ends may be varied at will, the nearer end 
having proportionally the greater influence on the resulting motion. 

A wheel rolling upon a plane is a case of aggregate motion, 
the center of the wheel moving parallel to the plane, and 
the wheel itself rotating upon its center. The resultant of 
these two motions gives the aggregate result of rolling. 

261. Pulley-blocks for 
Hoisting. The simple 
forms of hoisting-tackle, 
as in Fig. 385, are exam- 
ples of aggregate combi- 
nations. The sheaves C 
and D turn on a fixed axis, 
while A and B turn on a 
bearing from which the 
weight W is suspended. 
Fig. 386 is in effect the 
same as Fig. 385, but gives 
a clearer diagram for 
studying the linear velo- 
city ratio. Assume that 
j, 385 the bar ab with the sheaves 
A and B and the weight 
W has an upward velocity represented by v. 
sheave A, since the point c 




Fig. 386 



The effect of this at the 
at any instant is fixed, is equivalent 
to a wheel rolling on a plane, and there would be an upward linear 
velocity at d = 2 v. At the sheave B there is the aggregate motion due 



308 



ELEMENTS OF MECHANISM 



f//////////////WZT 



to the downward linear velocity at e = 2 v and the upward linear velocity 
of the axis b = v, giving for the linear velocity of /, 4 v upwards. 

mi j, linear speed F 4 W 

Therefore r - — -t-==. = 7 = -5- 

hnear speed W 1 F 

Many elevator-hoisting mechanisms are arranged in a similar man- 
ner, the force being applied at W, and the resulting force being given at F. 
This means a large force acting through a relatively small distance, 
producing a relatively small force acting through a much greater 
distance. 

The mechanical advantage of a hoist is the ratio of the weight 
which can be lifted to the force which is exerted, friction being 
neglected. 

The mechanical advantage of a given hoist can be determined by 
finding the velocity ratio as above and then, since the distances moved 
through in a given time (assuming constant velocity ratio) are directly 
as the velocities, the forces must be inversely as the velocities. Other 
methods of determining the mechanical 
advantage are illustrated by the fol- 
lowing examples.* 

Example 73. Hoist with Two Single Sheave 
Blocks. In Fig. 387 the upper block A, 
known as the standing block, is suspended 
from a fixed support. The rope is made fast 
to the casing of the upper block, passes around 
the sheave in the lower block and up around 
the sheave P which turns about the axis S in 
the upper block. It is required to find the 
force at F necessary to raise a weight W of 100 
lbs. suspended from the lower block. 

Solution. Assume that W is lifted 1 ft. by- 
some external force with the rope at F not 
moving. Then 1 ft. of slack rope would result 
at R and another foot of slack at T, giving a 
total of 2 ft. of slack which must be drawn 
over to F in order to keep the rope tight. 
Therefore, the linear speed of F is to the linear 
speed of IF as 2 is to 1. Hence, F is to W as 
1 is to 2, or F = \ W = 50 lbs. 

Example 74. Hoist with One Single Block 
and One Double Block. The hoist shown in 
Fig. 388 has the part of the rope which is 
marked T made fast to the lower block; it 
then passes over a sheave in the upper block, comes down at R and passes under the 
sheave in the lower block, then up at P over a second sheave in the upper block 
and off at F. 

* These solutions assume that the ropes are parallel. 



w 



Fig. 387 



W 



Fig. 388 



MISCELLANEOUS MECHANISMS 



309 



It is required to find the mechanical advantage of this hoist; that is, the ratio 
of the weight W to the force at F. 

Solution. Applying the same method used in Example 47, shows 1 ft. of slack in 
each of the three parts R, T, and P, or a total of 3 ft. which must be drawn off at 
F if W is lifted 1 ft. by an external force. Therefore, 



Example 75. "Luff on Luff." Fig. 389 shows a combination of two sets of 
pulley blocks, the rope F of the first set being made fast to the moving block of the 
second set. 

Solution. The mechanical advantage of each set is found as in the previous ex- 
amples. Then the product of the two is the mechanical advantage of the combina- 
tion. The first set in this case has a mechanical 
advantage of 3, and the second set of 4; therefore 
the combination has a mechanical advantage of 12. 
If the hook H were attached to a stationary support 
and the load applied to the hook K, the advantage 
r \\\ of the system would be 16. 



///^/SV7//////>. 





Fig. 389 



Fig. 390 



Example 76. "Spanish Burton." If the weight W (Fig. 390) is lifted 1 ft., a 
foot of slack is caused at both P and R. The foot at P is carried over to T which, 
in turn, causes a foot of slack in both R and F; this makes a total of 2 ft. of slack 
in R which must be drawn over to F in addition to the 1 ft. already given to F from T. 
Therefore, 3 ft. must be taken up at F for every foot that W is lifted. Then the 
mechanical advantage is 3. 



310 



ELEMENTS OF MECHANISM 



262. Weston Differential Pulley Block. Fig. 391 shows a chain 
hoist known as the Weston Differential Pulley Block. The two upper 
sheaves A and B are fast to each other. The diameter of A is a little 
larger than the diameter of B and it is the ratio of these two diameters 
which governs the mechanical advantage. 

The diameter of the lower sheave C is usually a mean between the 

diameters of the upper ones in order that the supporting chain may 

hang vertically. This feature is not of great importance, and the 

diameter of the lower sheave has no effect on the 

mechanical advantage. 

The chain is endless, passing over A, down at R, 
under C, up at P, around B, and hanging loose. 
The lifting force is applied at F. The sheaves are 
so shaped that the links of the chain fit into spaces 
provided for them to prevent slipping. 

The operation of the hoist may be seen from the 
following: 

Let D a represent the pitch diameter of the 
sheave A, D b the pitch diameter of the sheave B. 
Assume that the chain is drawn down at F fast 
enough to cause A to make one complete turn in 
a unit of time; that is, F has a speed of xD linear 
units. This would give an upward speed to the 
chain at R of irD a linear units. . Then, if B were not 
turning the sheave C would roll up on P, its 
center rising at a speed equal to one-half the 
speed of the chain at R; that is, the center of the lower sheave, 

and therefore the weight W, would rise at a speed of —^ linear 

units. But at the same time that R is rolling C up on P the pulley B is 
turning at the same angular speed as A, and therefore paying out chain 
at P at the rate of tDj, linear units per unit of time. This causes C 
to roll down on R at a speed such that its center is lowered at a speed of 

-pr- 6 linear units. The resultant upward speed of the center of C is, 
2i 

therefore, 

TrDa _ 7rD 6 = 7T (Dg - D b ) 

2 2 2 

Since the speed of F is ivD a the ratio of the speed of F to that of W is 

irDg 2Dg ■ 

ir{D a -D h ) ~ D a -D b ' 




Fig. 391 



MISCELLANEOUS MECHANISMS 



311 



The speed ratio may be found graphically as shown in Fig. 392. 
From E lay off along the chain a distance V representing the velocity 

of E. Draw a line (shown dotted) from the end of this distance, to 

the center of the sheave. The length 

Vi intercepted on the line of the 

chain through E\ is the velocity of 

E\. Draw V\ downward at the left- 
hand side of the lower sheave and V 

upward at the right-hand side of the 

same sheave. Join the ends of these 

two lines as shown, getting F 4 , the 

resultant velocity of M. The figure 

also shows, at Y% and V 3 , the effects 

of V and Vi respectively, when as- 
sumed to act successively. 

263. Intermittent Motion from 

Reciprocating Motion. A recipro- 
cating motion in one piece may 

cause an intermittent circular or 

rectilinear motion in another piece. 

It may be arranged that one-half 

of the reciprocating movement is 

suppressed and that the other half 

always produces motion in the same 

direction, giving the ratchet-wheel; 

or the reciprocating piece may act on opposite sides of a toothed wheel 
alternately, and allow the teeth to pass 
one at a time for each half reciprocation, 
giving the different forms of escapements 
as applied in timepieces. 

264. Ratchet-wheel. A wheel, pro- 
vided with suitably shaped pins or teeth, 
receiving an intermittent circular motion 
from some vibrating or reciprocating 
piece, is called a ratchet-wheel. 

In Fig. 393, A represents the ratchet- 
wheel turning upon the shaft a; C is an 

oscillating lever carrying the [detent, click, or catch B, which acts on 

the teeth of the wheel. The whole forms the three-bar linkage acb. 

When the arm C moves left-handed, the click B will push the 

wheel A before it through a space dependent upon the motion of 

C. When the arm moves back, the click will slide over the points 





312 



ELEMENTS OF MECHANISM 




Fig. 394 



of the teeth, and will be ready to push the wheel on its forward 
motion as before; in any case, the click is held against the wheel 
either by its weight or the action of a spring. In order that the 
arm C may produce motion in the wheel 
A, its oscillation must be at least suffi- 
cient to cause the wheel to advance one 
tooth. 

It is often the case that the wheel A 
must be prevented from moving back- 
ward on the return of the click B. In 
such a case a fixed pawl, click, or detent, 
similar to B, turning o na fixed pin, is ar- 
ranged to bear on the wheel, it being 
held in place by its weight or a spring. Fig. 393 might be taken to 
represent a retaining-pawl, in which case ac is a fixed link and the 
click B would prevent any right handed motion of the wheel A. 
Fig. 394 shows a retaining-pawl which would prevent rotation of the 
wheel A in either direction; such pawls are often used to retain pieces 
in definite adjusted positions. 

If the diameter of the wheel A (Fig. 393) be increased indefinitely, 
it will become a rack which would then receive an intermittent trans- 
lation on the vibration of the arm C; a retaining-pawl might be required 
in this case also to prevent a backward motion of the rack. 

A click may be arranged to push, as in Fig. 393, or to pull, as in Fig. 
400. In order that a click or pawl may retain its hold on the tooth of a 
ratchet-wheel, the common normal 
to the acting surfaces of the click 
and tooth, or pawl and tooth, must 
pass inside of the axis of a pushing 
click or pawl, as shown on the 
lowest click, Fig. 395, and outside the 
axis of the pulling click or pawl; 
the normal might pass through the 
axis, but the pawl would be more 
securely held if the normal is located 
according to the above rule, which 
also secures the easy falling of the 
pawl over the points of the teeth. 
It is sometimes necessary, or more convenient, to place the click- 
actuating lever on an axis different from that of the ratchet-wheel; in 
such a case care must be taken that in all positions of the click the 
common normal occupies the proper position; it will generally be suffi- 




MISCELLANEOUS MECHANISMS 313 

cient to consider only the extreme positions of the pawl in any case. 
Since when the lever vibrates on the axis of the wheel, the common 
normal always makes the same angle with it in all positions, thus 
securing a good bearing of the pawl on the tooth, it is best to use this 
construction when practicable. 

The effective stroke of a click or pawl is the space through which the 
ratchet-wheel is driven for each forward stroke of the arm. The total 
stroke of the arm should exceed the effective stroke by an amount 
sufficient to allow the click to fall freely into place. 

A common example of the application of the click and ratchet-wheel 
may be seen in several forms of ratchet-drills used to drill metals by 
hand. As examples of the retaining-pawl and wheel we have capstans 
and windlasses, where it is applied to prevent the recoil of the drum or 
barrel, for which purpose it is also applied in clocks. 

It is sometimes desirable to hold a drum at shorter intervals than 
would correspond to the movement of one tooth of the ratchet-wheel; 
in such a case several equal pawls may be used. Fig. 395 shows a 
case where three pawls were used, all attached by pins c, &, d to the 
fixed piece C, and so proportioned that they come into action alter- 
nately. Thus, when the wheel A has moved an amount corresponding 
to one-third of a tooth, the pawl B\ will be in contact with the tooth h; 
after the next one-third movement, B 2 will be in contact with 6 2 ; then 
after the remaining one-third movement, B will come into contact with 
the tooth under b; and so on. This arrangement enables us to obtain 
a slight motion and at the same time use comparatively large and strong 
teeth on the wheel in place of small weak ones. The piece C might 
also be used as a driving arm, and the wheel could then be moved 
through a space less than that of a tooth. The three pawls might be 
made of different lengths and placed side by side on one pin, as c h in 
which case a wide wheel would be necessary: the number of pawls 
required would be fixed by the conditions in each case. 

265. Reversible Click or Pawl. The usual form of the teeth of a 
ratchet-wheel is that given in Fig. 395, which only admits of motion in 
one direction; but in feed mechanisms, such as those in use on shapers 
and planers, it is often necessary to toake use of a click and ratchet- 
wheel that will drive in either direction. Such an arrangement is shown 
in Fig. 396, where the wheel A has radial teeth, and the click, which is 
made symmetrical, can occupy either of the positions B or B', thus giv- 
ing to A a right- or a left-handed motion. In order that the click B 
may be held firmly against the ratchet-wheel A in all positions of the 
arm C, its pivot c, after passing through the arm, is provided with a small 
triangular piece (shown dotted); this piece turning with B has a flat- 



314 



ELEMENTS OF MECHANISM 



ended presser, always urged upward by a spring (also shown dotted) 
bearing against the lower angle opposite B, thus urging the click toward 
the wheel; a similar action takes place when the click is in the dotted 
position B' . When the click is placed in line with the arm C, it is held 
in position by the side of the triangle parallel to the face of the click; 
thus this simple contrivance serves to hold the 
click so as to drive in either direction, and also to 
retain it in position when thrown out of gear. 

Since for different classes of work a change in the 
"feed" is desired, the arrangement must be such 
that the motion of the ratchet-wheel A (Fig. 396), 
which produces the feed, .can be adjusted. This is 
often done by changing the swing of the arm C, 
which is usually actuated by a rod attached at its 
free end. The other end of the rod is attached to a 
vibrating lever which has a definite angular move- 
ment at the proper time for the feed to occur, and 
is provided with a T slot in which the pivot for the 
rod can be adjusted by means of a thumb-screw and nut. By varying 
the distance of the nut from the center of motion of the lever, the swing 
of the arm C can be regulated; to reverse the feed, it occurring in the 
same position as before, the click must be reversed and the nut moved 
to the other side of the center of swing of the lever. 

Figs. 397 and 398 show other methods of adjusting the motion of the 
ratchet-wheel. In Fig. 397, which shows a form of feed mechanism 
used by Sir J. Whitworth in his planing-machine, C is an arm carrying 
the click B, and swinging loosely on the shaft a fixed to the ratchet- 




Fig. 396 





Fig. 397 



Fig. 398 



wheel A. The wheel E, also turning loosely on the shaft a, and placed 
just behind the arm C, has a definite angular motion sufficient to pro- 
duce the coarsest feed desired; its concentric slot m is provided with 
two adjustable pins ee, held in place by nuts at their back ends, and 
enclosing the lever C, but not of sufficient length to reach the click B. 
When the pins are placed at the ends of the slot, no motion will occur in 



MISCELLANEOUS MECHANISMS 315 

the arm C; but when e and e are placed as near as possible to each other, 
confining the arm C between them, all of the motion of E will be given 
to the arm C, thus producing the greatest feed; any other positions of 
the pins will give motions between the above limits, and the adjust- 
ment may be made to suit each case. 

In Fig. 398, the stationary shaft a, made fast to the frame of the 
machine at m, carries the vibrating arm C, ratchet-wheel A, and ad- 
justable shield S; the two former turn loosely on the shaft, while the 
latter is made fast to it by means of a nut n, the hole in S being made 
smaller than that in A, to provide a shoulder against which S is held 
by the nut. The arm C carries a pawl B of a thickness equal to that of 
the wheel plus that of the shield S; the extreme positions of this pawl 
are shown by dotted lines at B' and B". The teeth of the wheel A 
may be made of such shape as to gear with another wheel operating the 
feed mechanism; or another wheel, gearing with the feed mechanism, 
might be made fast to the back of A, if more convenient: in the latter 
case, the arm C would be placed back of this second wheel. 

If we suppose the lever in its extreme left position, the click will be 
at B" resting upon the face of the shield S, which projects beyond the 
points of the teeth of A ; and in the right-handed motion of the lever the 
click will be carried by the shield S until it reaches the position B, where 
it will leave the shield and come in contact with the tooth 6, which it will 
push to b' in the remainder of the swing. In the backward swing of 
the lever the click will be drawn over the teeth of the wheel and face of 
the shield to the position B". In the position of the shield shown in the 
figure a feed corresponding to three teeth of the wheel A is produced; 
by turning the shield to the left one, two, or three teeth, a feed of four, 
five, or six teeth might be obtained ; while, by turning it to the right, the 
feed could be diminished, the shield S being usually made large enough to 
consume the entire swing of the arm C. This form of feed mechanism 
has often been used in slotting-machines, and in such cases, as well as in 
Figs. 397 and 398, the click is usually held to its work by gravity. 

266. Double-acting Click. This device consists of two clicks mak- 
ing alternate strokes, so as to produce a nearly continuous motion of 
the ratchet-wheel which they drive, that motion being intermittent 
only at the instant of reversal of the movement of the clicks. In Fig. 
399 the clicks act by pushing, and in Fig. 400 by pulling; the former 
arrangement is generally best adapted to cases where much strength is 
required, as in windlasses. 

Each single stroke of the click-arms cdc' (Fig. 399) advances the 
ratchet-wheel through one-half of its pitch or some multiple of its half- 
pitch. To make this evident, suppose that the double click is to advance 



316 



ELEMENTS OF MECHANISM 



the ratchet-wheel one tooth for each double stroke of the click-arms, 
the arms being shown in their mid-stroke position in the figure. Now 
when the click be is beginning its forward stroke, the click b'c' has just 
completed its forward stroke and is beginning its backward stroke; 
during the forward stroke of be the ratchet-wheel will be advanced one- 
half a tooth; the click b'c', being at the same time drawn back one- 
half a tooth, will fall into position ready to drive its tooth in the remain- 




ing single stroke of the click-arms, which are made equal in length. By 
the same reasoning it may be seen that the wheel can be moved ahead 
some whole number of teeth for each double stroke of the click-arms. 

In Fig. 399 let the axis 'a and dimensions of the ratchet-wheel be 
given, also its pitch circle BB, which is located half-way between the 
tips and roots of the teeth. Draw any convenient radius ab, and from 
it lay off the angle bae equal to the mean obliquity of action of the clicks, 

that is, the angle that the lines of 
action of the clicks at mid-stroke are 
to make with the tangent to the pitch 
circle through the points of action. 
. , On ae let fall the perpendicular be, 
and with the radius ae describe the 
circle CC: this is the base circle, to 
which the lines of action of the clicks 
should be tangent. Lay off the angle 
eaf equal to an odd number of times 
the half-pitch angle, and through the 
points e and /, on the base circle, 
draw two tangents cutting each other in h. Draw hd bisecting the angle 
at h, and choose any convenient point in it, as d, for the center of the 
rocking shaft, to carry the click-arms. From d let fall the perpendiculars 
dc and dc' on the tangents hec and fhc' respectively; then c and c' will 
be the positions of the click-pins, and dc and dc' the center lines of the 




Fig. 400 



MISCELLANEOUS MECHANISMS 317 

click-arms at mid-stroke. Let b and b' be the points where ce and c'f 
cut the pitch circle; then cb and c'b' will be the lengths of the clicks. 
The effective stroke of each click will be equal to half the pitch as meas- 
ured on the base circle CC (or some whole number of times this half- 
pitch), and the total stroke must be enough greater to make the clicks 
clear the teeth and drop well into place. 

In Fig. 400 the clicks pull instead of push, the obliquity of action is 
zero, and the base circle and pitch circle become one, the points b, e, 
and b', f (Fig. 399) becoming e and / (Fig. 400) . In all other respects 
the construction is the same as when the clicks act by pushing, and the 
different points are lettered the same as in Fig. 399. 
^ Since springs are liable to lose their elasticity or become broken after 
being in use some time, it is often desirable to get along without apply- 
ing them to keep clicks in position. Fig. 401 shows in elevation a 
mechanism where no springs are required to keep the clicks in place, it 




Fig. 401 Fig. 402 

being used in some forms of lawn-mowers to connect the wheels to the 
revolving cutter when the mower is pushed forward, and to allow a 
free backward motion of the mower while the cutter still revolves. The 
ratchet A is usually made on the inside of the wheels carrying the mower, 
and the piece C, turning on the same axis as A, carries the three equi- 
distant pawls or clicks B, shaped to move in the cavities provided for 
them. In any position of C, at least one of the clicks will be held in 
contact with A by the action of gravity, and any motion of A in the 
direction of the arrow will be given to the piece C. Here the ratchet- 
wheel drives the click, ac being the actuated click-lever. The piece C 
is sometimes attached to a roller by means of the shaft a; then any 
left-handed motion of C will be given to A, while the right-handed 
motion will simply cause the clicks to slide over the teeth of A. The 
clicks B are usually held in place by a cap attached to C. 
Fig. 402 shows a form of click which is always thrown into action 



318 



ELEMENTS OF MECHANISM 



when a left-handed rotation is given to its arm C, while any motion of 
the wheel A left-handed will immediately throw the click out of action. 
The wheel A carries a projecting hub d, over which a spring D is fitted 
so as to move with slight friction. One end of this spring passes between 
two pins, e, placed upon an arm attached to the click B. When the 
arm C is turned left-handed, the wheel A and the spring D being sta- 
tionary, the click B will be thrown toward the wheel by the action of 
the spring on the pin e. The motion of the wheel A will be equal to 
that of the arm C, minus the motion of C necessary to throw the click 
into gear. Similarly, when A turns left-handed, the click B is thrown 
out of gear. This mechanism is employed in some forms of spinning- 
mules to actuate the spindles when winding on the spun yarn. 

267. Friction-catch. Various forms of catches depending upon 
friction are often used in place of clicks ; these catches usually act upon 
the face of the wheel or in a suitably formed groove cut in the face. 
Friction-catches have the advantage of being noiseless and allowing any 
motion of the wheel, as they can take hold at any point; they have the 
disadvantage, however, of slipping when worn, and of getting out of 
order. 

Fig. 403 shows a friction-catch B working in a V-shaped groove in 
the wheel A, as shown in section A'B'. Here B acts as a retaining 

I 





Fig. 403 



Fig. 404 



click, and prevents any right-handed motion of A ; its face is circular 
in outline, the center being located at d, a little above the axis c. A 
similarly shaped catch might be used in place of an actuating click to 
cause motion of A. 

Fig. 404 shows four catches like B (Fig. 403) applied to drive an 
annular ring A in the direction indicated by the arrow. When the 
piece c is turned right-handed, the catches B are thrown against the 
inside b of the annular ring by means of the four springs shown; on 
stopping the motion of c, the pieces B are pushed, by the action of h, 



MISCELLANEOUS MECHANISMS 



319 



toward the springs which slightly press them against the ring and 
hold them in readiness to again grip when c moves right-handed. Thus 
an oscillation of the piece c might cause continuous rotation of the 
wheel A, provided a fly-wheel were applied to A to keep it going while 
c was being moved back. The annular ring A is fast to a disk carried 
by the shaft a; the piece c turning loosely on a has a collar to keep it 
in position lengthwise of the shaft. 

The nipping-lever shown in Fig. 405 is another application of the 
friction-catch. A loose ring C surrounds the wheel A ; a friction-catch 
B having a hollow face works in a pocket in the ring and is pivoted 
at c. On applying a force at the end of the catch B in the direction 
of the arrow, the hollow face of the catch will "nip" the wheel at b, 
and cause the ring to bear tightly against the left-hand part of the cir- 
cumference of the wheel; the friction thus set up will cause the catch, 
ring, and wheel to move together as one piece. The greater the pull 
applied at the end of the catch the greater will be the friction, as the 
friction is proportional to the pressure; thus the amount of friction 
developed will depend upon the resistance to motion of A. Upon re- 
versing the force at the end of the catch, the hollow face of the catch 
will be drawn away from the face of A, and the rounding top part 
of the catch, coming in contact with the top of the cavity in the ring, 
will cause the ring to slide back upon the disk. An upward motion of 
the click end will again cause the wheel A to move forward, and thus 
the action is the same as in a ratchet and wheel. 

Fig. 406 shows, in section, a device which has been applied to actu- 
ate sewing-machines in place of the common crank. Two such mechan- 





Fig. 405 



Fig. 406 



isms were used, one to rotate tne shaft of the machine on a downward 
tip of the treadle, while the other acted during the upward tip, the 
treadle-rods being attached to the projections of the pieces B. The 
mechanism shown in the figure acts upon the shaft during the down- 
ward motion of the projection B as shown by the arrow. 

The piece C, containing an annular groove, is made fast to the shaft a, 
the sides of this groove being turned circular and concentric with the 



320 



ELEMENTS OF MECHANISM 



shaft. The piece B, having a projecting hub fitting loosely on the 
inner surface of the groove in C, is placed over the open groove, and 
is held in place by a collar on the shaft. The hub on the piece B, and 
the piece C, are shown in section. The friction-catch D, working in the 
groove, is fitted over the hub of B, the hole in D being elongated in 
the direction ab so that D can move slightly upon the hub and between 
the two pins e fast in the piece B. A cylindrical roller c is placed in the 
wedge-shaped space between the outer side of the groove and the piece 
D, a spring always actuating this roller in a direction opposite to that 
of the arrow, or towards the narrower part of the space. 

Now when the piece B is turned in the direction of the arrow by a 
downward stroke of the treadle-rod, it will move the piece D with it by 
means of the pins e; at the same time, the roller c will move into the 
narrow part of the wedge-shaped space between C and D, and cause 
binding between the pieces D and C at b and at the surface of the roller. 
The friction at b thus set up will cause the motion of D to be given to 
C. On the upward motion of the projection B the roller will be moved 
to the large part of its space by the action of the piece C revolving with 
the shaft combined with that of the backward movement of D, thus 
releasing the pressure at b and allowing C to move freely onward. 
The other catch would be made just the reverse of this one, and would 
act on amupward movement of the treadle-rod. 

Another form of friction catch, some- 
times used in gang saws to secure the 
advance of the timber for each stroke of 
the saw, and called the silent feed, is shown 
in Fig. 407. 

The saddle-block B, which rests upon 
the outer rim of the annular wheel \A, 
carries the lever C turning upon the pin c. 
The block D, which fits the inner rim of 
the wheel, is carried by the lever C, and 
is securely held to its lower end by the pin 
d on which D can freely turn. When the pieces occupy the positions 
shown in the figure, a small space exists between the piece D and the 
inside of the rim A. 

The upper end of the lever C has a reciprocating motion imparted to 
it by means of the rod E. The oscillation of the lever about the pin c 
is limited by the stops e and G carried by the saddle-block B. When 
the rod E is moved in the direction indicated by the arrow, the lever 
turning on c will cause the block D to approach B, and thus nip the rim 
at a and b; and any further motion of C will be given to the wheel A. 




Fig. 407 



i 



MISCELLANEOUS MECHANISMS 321 

On moving E in the opposite direction the grip will first be loosened, 
and the lever striking against the stop e will cause the combination to 
slide freely back on the rim A . The amount of movement given to the 
wheel can be regulated by changing the stroke of the rod E by an 
arrangement similar to that described in connection with the reversible 
click, § 265. The stop G can be adjusted by means of the screw F so 
as to prevent the oscillation of the lever upon its center c, thus throw- 
ing the grip out of action. The saddle-block B then merely slides 
back and forth on the rim, the action being the same as that obtained 
by throwing the ordinary click out of gear. 

268. Masked Wheels. It is sometimes required that certain 
strokes of the click-actuating lever shall remain inoperative upon the 
ratchet-wheel. Such arrangements are made use of in numbering- 
machines where it is desired to print the same number twice in succes- 
sion; they are called masked wheels. 

Fig. 408, taken from a model, illustrates the action of a masked 
wheel; the pin-wheel D represents the first ratchet-wheel, and is fast 
to the axis a; the second wheel A has its teeth 
arranged in pairs, every alternate tooth being 
cut deeper, and it turns loosely on the axis a. 
The click B is so made that one of its acting 
surfaces, i, bears against the pins e of the wheel 
D, while the other, g, is placed so as to clear the 
pins and yet bear upon the teeth of A, the wheel 
A being located so as to admit of this. 

If now we suppose the lever C to vibrate 
through an angle sufficient to move either wheel 
along one tooth, both having the same number, 
it will be noticed that when the projecting piece g 

is resting in a shallow tooth of the wheel A, the acting surface i will be 
retained too far from the axis to act upon the tooth e, and thus this 
vibration of the lever will have no effect upon the pin- wheel D; while 
when the piece g rests in a deep tooth, as b', the click will be allowed 
to drop so as to bring the surface i into action with the pin e'. 

In the figure the click B has just pushed the tooth e' into its present 
position, the projection g having rested in the deep tooth b' of the wheel 
A ; on moving back, g has slipped into the shallow tooth 6, and thus the 
next stroke of the lever and click will remain inoperative on the wheel 
D, which advances but one tooth for every two complete oscillations of 
the lever C. 

Both wheels should be provided with retaining-pawls, one of which, 
p, is shown. This form of pawl, consisting of a roller p turning about 




322 



ELEMENTS OF MECHANISM 



a pivot carried by the spring s, attached to the frame carrying the mech- 
anism, is often used in connection with pin-wheels, as by rolling between 
the teeth it always retains them in the same position relative to the axis 
of the roller; a triangular-pointed pawl which also passes between the 
pins is sometimes used in place of the roller. 

The pins of the wheel D might be replaced by teeth so made that 
their points would be just inside of the bottoms of the shallow teeth of 
A; a wide pawl would then be used, and when it rested in a shallow 
tooth of A it would remain inoperative on D, while when it rested in a 
deep tooth it would come in contact with the adjacent tooth of D and 
push it along. 

So long as the click B and the wheels have the proper relative motion 
it makes no difference which we consider as fixed, as the action will be 
the same whether we consider the axis of the wheels as fixed and the 
click to move, or the click to be fixed and the axis to have the proper 
relative motion in regard to it. The latter method is made use of in 
some forms of numbering-machines. 

269. Counter Mechanism. Fig. 409 shows the mechanism of a 
"counter" used to record the number of double strokes made by a 
pump; the revolutions made by a steam engine, paddle, propeller, or 
other shaft, etc. Two views are given in the figure, which represents 
a counter capable of recording revolutions from 1 
to 999; if it is desired to record higher numbers, it 
will only be necessary to add more wheels, such as 
A. A plate, having a long slot or series of openings 
opposite the figures 000, is placed over the wheels, 
thus only allowing the numbers to be visible as they 
come under the slot or openings. 

The number wheels A, A h A 2 are arranged to 
turn loosely side by side upon the small shaft a, 
and are provided with a series of ten teeth cut into 
one side of their faces, while upon the other side a 
single notch is cut opposite the zero tooth on the 
first side, it having the same depth and contour. 
This single notch can be omitted on the last wheel A 2 . The numbers 
0, 1, 2, 3, 4, 5, 6, 7, 8, 9 are printed upon the faces of the wheels in 
proper relative positions to the teeth t. 

Two arms C are arranged to vibrate upon the shaft a of the number 
wheels, and carry at their outer ends the pin c, on winch a series of 
clicks, b, 6i, and b 2) are arranged, collars placed between them serving 
to keep them in position on the pin. The arms are made to vibrate 
through an angle sufficient to advance the wheels one tooth, i.e., one- 




Fig. 409 



MISCELLANEOUS MECHANISMS 323 

tenth of a turn; their position after advancing a tooth is shown by 
dotted lines in the side view. A common method of obtaining this 
vibration is to attach a rod at r, one end of the pin c, this rod to be so 
attached at its other end to the machine as to cause the required back- 
ward and forward vibrations of the lever C for each double stroke or 
revolution that the counter is to record. 

The click 6 is narrow, and works upon the toothed edge of the first 
wheel A, advancing it one tooth for every double stroke of the arm c. 
The remaining clicks 61 and b 2 are made broad, and work on the toothed 
edges of Ai and A 2 , as well as on the notched rims of A and A h re- 
spectively. When the notches n and m come under the clicks 61 and b 2 
the clicks will be allowed to fall and act on the toothed parts of Ai and 
A2) but in any other positions of the notches the clicks will remain 
inoperative upon the wheels, simply riding upon the smooth rims of A 
and Ax, which keep the clicks out of action. Each wheel is provided 
with a retaining-spring s to keep it in proper position. 

Having placed the wheels in the position shown in the figure, the 
reading being 000, the action is as follows: The click b moves the 
wheel A along one tooth for each double stroke of the arm C, the clicks 

61 and b 2 remaining inoperative on Ai and A 2 ; on the figure 9 reaching 
the slot, or the position now occupied by 0, the notch n will allow the 
click 61 to fall into the tooth 1 of the wheel Ai, and the next forward 
stroke of the arm will advance both the wheels A and A\, giving the 
reading 10; the notch n having now moved along, the click 61 will 
remain inoperative until the reading is 19, when 61 will again come into 
action and advance A\ one tooth, giving the reading 20; and so on up 
to 90, when the notch m comes under the click b 2 . To prevent the click 

62 from acting on the next forward stroke of the arm, which would 
make the reading 101 instead of 91, as it should be, a small strip i is 
fastened firmly to the end of the click b 2 , its free end resting upon the 
click 61. This strip prevents the click b 2 from acting until the click &i 
falls, which occurs when the reading is 99; on the next forward stroke 
the clicks 61 and 62 act, thus giving the reading 100. As the strip merely 
rests upon 61, it cannot prevent its action at any time. If another 
wheel were added, its click would require a strip resting on the end of 
62. A substitute for these strips might be obtained by making the 
wheel A fast to the shaft a, and allowing the remaining wheels to turn 
loose upon it, thinMisks, having the same contour as the notched edge of 
the wheel A, being placed between the wheels A X A 2 , A 2 A 3 , etc., and made 
fast to the shaft, the notches all being placed opposite n; thus the edges 
of the disks would keep the clicks b 2 , 6 3 , etc., out of action, except when 
the figure 9 of the wheel A is opposite the slot, and the notches m, etc., 



324 



ELEMENTS OF MECHANISM 



are in proper position. A simpler form of counter will be described in 
§270. 

270. Intermittent Motion from Continuous Motion. The cases of 
intermittent motion thus far considered have been those in which a 
uniform reciprocating motion in one piece gives an intermittent cir- 
cular or rectilinear motion to another, the click being the driver and 
the wheel the follower. 

It is often required that a uniform circular motion of the driver 
shall produce an intermittent circular or rectilinear mdtion of the fol- 
lower. The following examples will give some solutions of the problem: 
Fig. 410 shows a combination by which the toothed wheel A is 
moved in the direction of the arrow, one tooth for every complete 
turn of the shaft d, the pawl B retaining the wheel in position when the 
tooth t on the shaft d is out of action. The stationary link adc forms 
the frame, and provides bearings for the shafts d and a, and a pin c for 

the pawl B. The arm e, placed 
by the side of the tooth upon the 
shaft, is arranged to clear the 
wheel A in its motion, and to lift 
the pawl B at the time when the 
tooth t comes into action wth the 
wheel, and to drop the pawl when 
the action of t ceases, i.e., when 
the wheel has been advanced one 




Fig. 410 



tooth. This is accomplished by attaching the piece n to the pawl, its 
contour in the raised position of the pawl being an arc of a circle about 
the center of the shaft d; its length is arranged to suit the above re- 
quirements. When the tooth t comes in contact with the wheel, the 
arm e, striking the piece n, raises the pawl (which is held in position by 
the spring s), and retains it in the raised position until the tooth t is 
ready to leave the wheel, when e, passing off from the 
end of n, allows the pawl to drop. 

In Fig. 411 the wheel A makes one-third of a revolu- 
tion for every turn of the wheel be, its period of rest 
being about one-half the period of revolution of be. 
If we suppose A the follower, and to turn right-handed 
while the driver be turns left-handed, one of the round 
pins b is just about to push ahead the long tooth of A, 
the circular retaining sector c being in such a position as to follow a 
right-handed motion of A. The first pin slides down the long tooth, and 
the other pins pass into and gear with the teeth b', the last pin passing 
off on the long tooth e, when the sector c will come in contact with the 




Fig. 411 



MISCELLANEOUS MECHANISMS 



325 




arc c', and retain the wheel A until the wheel be again reaches its 
present position. 

Fig. 412 is a diagram of a mechanism known as a Geneva stop. 
The wheel A makes one-sixth of a revolution for one turn of the driver 
ac, the pin b working in the slots b' causing the motion of A ; while the 
circular portion c of the driver, coming in contact with 
the corresponding circular hollows c' , retains A in position 
when the tooth b is out of action. The wheel a is cut 
away just back of the pin b to provide clearance for the 
wheel A in its motion. If we close up one of the slots, 
as b', it will be found that the shaft a can only make a 
little over five and one-half revolutions in either direction 
before the pin b will strike the closed slot. This 
mechanism, when so modified, has been applied to 
watches to prevent overwinding, and is called the Geneva stop, the 
wheel a being attached to the spring-shaft so as to turn with it, while 
A turns on an axis d in the spring-barrel. The number of slots in A 
depends upon the number of times it is desired to turn the spring- 
shaft. 

By placing another pin opposite b in the wheel ac, as shown by dotted 
lines, and providing the necessary clearance, the wheel A could be 
moved through one-sixth of a turn for every half turn of ac. 

A simple type of counter extensively used on water-meters is shown 
in Fig. 413. It consists of a series of wheels A, B, C, mounted side by 



Fig. 412 




Fig. 413 



side and turning loosely on the shaft S; or the first wheel to the right 
may be fast to the shaft and all the remaining wheels loose upon it. 
Each wheel is numbered on its face as in Fig. 409, and it is provided, as 
shown, that the middle row of figures appears in a suitable slot in the 
face of the counter. The first wheel A is attached to the worm-wheel E, 



326 



ELEMENTS OF MECHANISM 



having 20 teeth and driven by the worm F geared to turn twice for one 
turn of the counter driving shaft. 

On a parallel shaft T loose pinions D are arranged between each 
pair of wheels. Each pinion is supplied with six teeth on its left side 
extending over a little more than one-half its face and with three teeth, 
each alternate tooth being cut away, for the remainder of the face, as 
clearly shown in the sectional elevations. The middle elevation (Fig. 
413) shows a view of the wheel B from the right of the line ah with the 
pinion D sectioned on the line cd. The right elevation shows a view of 
the wheel A from the left of the line ab with the pinion D sectioned on 
the line cd. The first wheel A, and all others except the last, at the 
left, have on their left sides a double tooth G, which is arranged to come 
in contact with the six-tooth portion of the pinion; the space between 
these teeth is extended through the brass plate which forms the left 
side of the number ring whose ^periphery H acts as a stop for the three- 
tooth portion of the pinion, as clearly shown in the figure to the right. 
Similarly on the right side of each wheel, except the first, is placed a 
wheel of 20 teeth gearing with the six-tooth part of the pinion, as shown 
in the middle figure. When the digit 9 on any wheel, except the one at 
the left, comes under the slot, the double tooth G is ready to come in 
contact with the pinion; as the digit 9 passes under the slot the tooth 
G starts the pinion, which is then free to make one-third of a turn and 
again become locked by the periphery H. Thus any wheel to the left 
receives one-tenth of a turn for every passage of the digit 9 on the wheel 
to its right. In the figure the reading 329 will change to 330 on the 
passing of the digit 9. This counter can be made to record oscillations 
by supplying its actuating shaft with a ten-tooth ratchet, arranged with 
a click to move one tooth for each double oscillation. 

Figs. 414a and 414b show two methods of advancing the wheels A 
through a space corresponding to one tooth during a small part of a 
revolution of the shafts c; in this case the shafts are at right angles to 
each other. In Fig. 414a a raised circular ring with a small spiral 





Fig. 414a 



Fig. 414b 



part b attached to a disk is made use of; the circular part of the ring 
retains the wheel in position, while the spiral part gives it its motion. 
In Fig. 414b the disk carried by the shaft cc has a part of its edge bent 



MISCELLANEOUS MECHANISMS 



327 



helically at b; this helical part gives motion to the wheel, and the 
remaining part of the disk edge retains the wheel in position. By- 
using a regular spiral, in Fig. 414a, and one turn of a helix, in Fig. 
414b, the wheels A could be made to move uniformly through the 
space of one tooth during a uniform revolution of the shafts c. 

In Fig. 415 the wheel A is arranged to turn the wheel B, on a shaft 
at right angles to that of A, through one-half a turn while it turns one- 
sixth of a turn, and to 
lock B during the remain- 
ing five-sixths of the turn. 
Fig. 416 illustrates the 
Star Wheel. The wheel 
A, turns through a space 
corresponding to one 
tooth for each revolution 
of the arm carrying the 
pin b and turn- 
ing on the shaft 
c. The pin b is 
often stationary, 
and the star 
wheel is moved 

Fig. 415 P ast lt ' the FiG. 416 

action is then 

evidently the same, as the pin and wheel have the same relative motion 
in regard to each other during the time of action. The star wheel is 
often used on moving parts of machines to actuate some feed mechan- 
ism, as may be seen in cylinder- 
boring machines on the facing 
attachment, and in spinning- 
machinery. 

271. Cam and Slotted 
Sliding Bar. Fig. 417 shows 
an equilateral triangle abc, 
formed by three circular arcs, 
whose centers are at a, b, and c, 
the whole turning about the 
axis a, and producing an inter- 
mittent motion in the slotted 





Fig. 417 



Fig. 418 



piece B. The width of the slot is equal to the radius of the three circular 
arcs composing the three equal sides of the triangular cam A , and there- 
fore the cam will always bear against both sides of the groove. 



328 ELEMENTS OF MECHANISM 

If we imagine the cam to start from the position shown in Fig. 418 
when 6 is at 1, the slotted piece B will remain at rest while b moves 
from 1 to 2 (one-sixth of the circle 1, 2 . . . 6), the cam edge be merely 
sliding over the lower side of the slot. When b moves from 2 to 3, i.e., 
from the position of A, shown by light full lines, to that shown by dotted 
lines, the edge ab will act upon the upper side of the slot, and impart 
to B a motion similar to that obtained in Fig. 357, being that of a 
crank with an infinite connecting-rod; from 3 to 4 the point 6 will drive 
the upper side of the slot, ca sliding over the lower side, the motion here 
being also that of a connecting-rod with an infinite link, but decreasing 
instead of increasing as from 2 to 3. When b moves from 4 to 5 there 
is no motion in B; from 5 to 6, c acts upon the upper side of the slot, 
and B moves downward; from 6 to 1, ac acts on the upper side of the slot, 
and B moves downward to its starting position. The motion of B is 
accelerated from 5 to 6 and retarded from 6 to 1. 

At A' a form of cam is shown where the shaft a is wholly contained 
in the cam. In this case draw the arcs de and cb from the axis of the 
shaft as a center, making ce equal to the width of the slot in B; from c 
as a center with a radius ce draw the arc eb, and note the point b where 
it cuts the arc cb; with the same radius and b as a center draw the arc 
dc, which will complete the cam. In this case the angle cab will not be 
equal to 60°, and the motions in their durations and extent will vary a 
little from those described above. 

272. Locking Devices. The principle of the slotted sliding bar 
combined with that of the Geneva stop is applied in the shipper mechan- 
ism shown in Fig. 419, often used on machines where the motion is 
automatically reversed. The shipper bar B 
slides in the piece CC, which also provides a 
pivot a for the weighted lever wab. The end 
of the lever b opposite the weight w carries a 
p. 419 pin which works in the grooved lug s on the 

shipper bar. In the present position of the 
pieces, the pin b is in the upper part of the slot, and the weight w, tending 
to fall under the action of gravity ,fholds it there, the shipper being thus 
effectually locked in its present position. If now the lever be turned left- 
handed about its'axis a until the weight w is just a little to the left of a, 
gravity will carry the weight and lever into the dotted position shown, 
where it will be locked until the lever is turned right-handed. The princi- 
ple of using a weight to complete the motion is very convenient, as the part 
of the machine actuating the shipper often stops before the belt is 
carried to the wheel which produces the reverse motion, and the machine 
is thus stopped. The motion can always be made sufficient to raise a 




MISCELLANEOUS MECHANISMS 329 

weighted lever, as shown above, and the weight will, in falling, com- 
plete the motion of the shipper. 

The device shown in Fig. 420, of which there may be many forms, 
serves to retain a wheel A in definite adjusted positions, its use being 
the same as that of the retaining-pawl shown in Fig. 394. The wheels 
B and A turn on the shafts c and a, respectively, carried by the link C, 
which is shown dotted, as it has been cut away in taking the section. 
Two positions of the wheel B will allow the teeth 6 of A to pass freely 
through its slotted opening, while any other position effectually locks 
the wheel A. The shape of the slot in B and the teeth of A are clearly 
shown in the figure. 

Fig. 421 shows another device for locking the wheel A, the teeth of 
which are round pins; but in this case it is necessary to turn B once to 
pass a tooth of A. If we suppose the wheel A under the influence of 





Fig. 420 Fig. 421 

a spring which tends to turn it right-handed, and then turn B uniformly 
either right- or left-handed, the wheel A will advance one tooth for each 
complete turn of B, a pin first slipping into the groove on the left and 
leaving it when the groove opens toward the right, the next pin then 
coming against the circular part of B opposite the groove. It will be 
noticed that while there are only six pins on the wheel A, yet there are 
twelve positions in which A can be locked, as a tooth may be in the 
bottom of the groove or two teeth may be bearing against the circular 
outside of B. Devices similar in principle to those shown in Figs. 420 
and 421 are often used to adjust stops in connection with feed 
mechanisms. 

Clicks and pawls as used in practice may have many different forms 
and arrangements; their shape depends very much upon their strength 
and the space in which they are to be placed, and the arrangement 
depends on the requirements in each case. 




330 ELEMENTS OF MECHANISM 

273. Escapements. An escapement is a combination in which a 
toothed wheel acts upon two distinct pieces or pallets attached to a re- 
ciprocating frame, it being so arranged that when one tooth escapes or 
ceases to drive its pallet, another tooth shall begin its action on the 
other pallet. 

A simple form of escapement is shown in Fig. 422. The frame cd 
is arranged to slide longitudinally in the bearings CC, which are at- 
tached to the bearing for the toothed 
wheel. The wheel a turns continually 
in the direction of the arrow, and is 
provided with three teeth, b, b', b", the 
frame having two pallets, c and c'. In 
the position shown, the tooth b is just 
Fig. 422 ceasing to drive the pallet c to the right, 

and is escaping, while the tooth b' is 
just coming in contact with the pallet c', when it will drive the frame 
to the left. 

While escapements are generally used to convert circular into recip- 
rocating motion, as in the above example, the wheel being the driver, 
yet, in many cases, the action may be reversed. In Fig. 422, if we 
consider the frame to have a reciprocating motion and use it as the 
driver, the wheel will be made to turn in the opposite direction to that 
in which it would itself turn to produce reciprocating motion in the 
frame. It will be noticed also that there is a short interval at the 
beginning of each stroke of the frame in which no motion will be given 
to the wheel. It is clear that the wheel a must have 1, 3, 5, or some 
odd number of teeth upon its circumference. 

274. The Crown-wheel Escapement. The crown-wheel escape- 
ment (Fig. 423) is used for causing a vibration in one axis by means of 
a rotation of another. The latter carries a crown 
wheel A, consisting of a circular band with an odd 
number of large teeth, like those of a splitting-saw, 
cut on its upper edge. The vibrating axis, o, or 
verge as it is often called, is located just above the 
teeth of the crown wheel, in a plane at right angles 
to the vertical wheel axis. The verge carries two 
pallets, b and b h located in planes passing through 
its axis, the distance between them being arranged 

so that they may engage alternately with teeth on opposite sides of 
the wheel. If the crown wheel be made to revolve under the action 
of a spring or weight, the alternate action of the teeth on the pallets 
will cause a reciprocating motion in the verge. The rapidity of this 




MISCELLANEOUS MECHANISMS 



331 



vibration depends upon the inertia of the verge, which may be adjusted 
by attaching to it a suitably weighted arm. 

This escapement, having the disadvantage of causing a recoil in the 
wheel as the vibrating arm cannot be suddenly stopped, is not used in 
timepieces, and but rarely in other places. It is of interest, however, 
as being the first contrivance used in a clock for measuring time. 

275. The Anchor Escapement. The anchor escapement as applied 
in clocks is shown in Fig. 424. The escape-wheel Ai turns in the direc- 
tion of the arrow and is supplied with long pointed teeth. The pallets 
are connected to the vibrating axis or verge C\ by means of the arms 
diCi and ei&, the axis of the verge and wheel being parallel to each 




Fig. 424 



other. The verge is supplied at its back - end with an arm &pi, carry- 
ing a pin pi at its lower end. This pin works in a slot in the pendulum- 
rod, not shown. The resemblance of the two pallet arms combined 
with the upright arm to an anchor gave rise to the name "anchor 
escapement." The left-hand pallet, di, is so shaped that all the nor- 
mals to its surface pass above the verge axis Ci, while all the normals 
to the right-hand pallet, e h pass below the axis C\. Thus an upward 
movement of either pallet will allow the wheel to turn in the direction 
of the arrow, or, the wheel turning in the direction of the arrow, will, 
when the tooth b x is in contact with the pallet d h cause a left-handed 
swing of the anchor; and when 6i has passed off from d\ and 0\ reaches 
the right-hand pallet, as shown, a right-handed swing will be given to 
the anchor. As the pendulum cannot be suddenly stopped after a 



332 ELEMENTS OF MECHANISM 

tooth has escaped from a pallet, the tooth that strikes the other pallet 
is subject to a slight recoil before it can move in the proper direction, 
which motion begins when the pendulum commences its return swing. 
The action of the escape-wheel on the pendulum is as follows : 

Suppose the points h and h to show extreme positions of the point pi, 
and suppose the pendulum and point pi to be moving to the left; the 
tooth 61 has just escaped from the pallet d h and Oi has impinged on ei,. 
as shown, the point pi having reached the position mi. The recoil 
now begins, the pallet ei moving back the tooth o h while pi goes from 
m\ to l\. The pendulum then swings to the right and the pallet e\ is 
urged upward by the tooth o h thus urging the pendulum to the right 
while pi passes from k to n\, when Oi escapes. Recoil then occurs on 
the pallet di from ni to k\, and from k\ to mi an impulse is given to the 
pendulum to the left, when the above-described cycle will be repeated. 
As the space through which the pendulum is urged on exceeds that 
through which it is held back, the action of the escape-wheel keeps the 
pendulum vibrating. This alternate action with and against the pen- 
dulum prevents it from being, as it should be, the sole regulator of the 
speed of revolution of the escape-wheel; for its own time of vibration, 
instead of depending only upon its length, will also depend upon the 
force urging the escape-wheel round. Therefore any change in the 
maintaining force will disturb the rate of the clock. 

276. Dead-beat Escapement. The objectionable feature of the 
anchor escapement is removed in Graham's dead-beat escapement, 
shown in Fig. 425. The improvement consists in making the outline of 
the lower surface, db, of the left-hand pallet, and the upper surface of 
the right-hand pallet, arcs of a circle about C, the verge axis; the 
oblique surfaces b and / complete the pallets. The construction indi- 
cated by dotted lines in the figure insures that the oblique surfaces of 
the pallets shall make equal angles, in their normal position, with the 
tangents bC and fC to the wheel circle not shown. If we suppose the 
limits of the swing of the point p to be I and k, the action of the escape- 
wheel on the pendulum is as follows : 

The pendulum being in its right extreme position, the tooth b is 
bearing against the circular portion of the pallet d; as the pendulum 
swings to the left under the action of gravity, the tooth b will begin to 
move along the inclined face of the pallet when the center line has 
reached n, and will urge the pendulum onward to m, where the tooth 
leaves the pallet, and another tooth o comes in contact with the cir- 
cular part of the pallet e, which, with the exception of a slight friction 
between it and the point of the tooth, will leave the pendulum free to 
move onward, the wheel being locked in position. On the return 



MISCELLANEOUS MECHANISMS 



333 



swing of the pendulum, the inclined part of the pallet e urges the pen- 
dulum from m to n. Hence there is no recoil, and the only action 
against the pendulum is the very minute friction between the teeth 
and the pallets. The impulse is here given through an arc mn, very 
nearly bisected by the middle point of the swing of the pendulum, 
which is also an advantage. The term "dead-beat" has been applied 




Fig. 425 

because the second hand, which is fitted to the escape-wheel, stops so 
completely when the tooth falls upon the circular portion of a pallet, 
there being no recoil or subsequent trembling such as occurs in other 
escapements. 

In watches the pendulum is replaced by a balance-wheel swinging 
backward and forward on an arbor under the action of a very light 
coiled spring, often called a " hair-spring " the pivots of the arbor are 
very nicely made, so that they turn with very slight friction. 

277. The Graham Cylinder Escapement. This form 
of escapement is used in the Geneva watches. Here the 
balance verge o (Fig. 426) has attached to it a very thin 
cyclindrical shell rs centered at o, the axis of the verge, and 
the point of the tooth b can rest either on the outside or 
inside of the cylinder during a part of the swing of the bal- 
ance. As the cylinder turns in the direction of the arrow 
(Fig. 426a), the wheel also being urged in the direction of 
its arrow, the inclined surface of the tooth be comes under 
the edge s of the cylinder, and thus urges the balance onward; this gives 
one impulse, as shown in Fig. 426b. The tooth then passes s, flies into 




Fig. 426b 



334 



ELEMENTS OF MECHANISM 



the cylinder, and is stopped by the concave surface near r. In the 
opposite swing of the balance the tooth escapes from the cylinder, the 
inclined surface pushing r upward, which gives the other impulse in the 
opposite direction to the first; the action is then repeated by the next 
tooth of the wheel. 

This escapement is, in its action, nearly identical to the dead-beat; 
but the impulse is here given through small equal arcs, situated at 
equal distances from the middle point of the swing. 

278. The Chronometer Escapement is shown in Fig. 427. Here 
the verge o carries two circular plates, one of which carries a projec- 
tion p, which serves to operate the detent d; the other carries a pro- 
jection n, which swings freely by the teeth of the escape-wheel when a 




Fig. 427 



tooth is resting upon the pallet d, but encounters a tooth when the 
wheel is in any other position. 

The detent d has a compound construction and consists of four 
parts : 

1° The locking-stone d, a piece of ruby on which the tooth of the 
escape-wheel rests. 

2° The discharging-spring Z, a very fine strip of hammered gold. 

3° A spring s on which the detent swings, and which attaches the 
whole to the frame of the chronometer. 

4° A support e, attached to the body of the detent, to prevent the 
strip I from bending upward. 

A pin r prevents the detent from approaching too near the wheel. 

The action of the escapement is as follows: On a right-hand swing 
of the balance the projection p meets the light strip I, which, bending 
from its point of attachment to the detent, offers but very little resist- 



MISCELLANEOUS MECHANISMS 335 

ance to the balance. On the return swing of the balance, the projec- 
tion p meets the strip I, which can now only bend from e, and raises 
the detent d from its support r, thus allowing the tooth 6 to escape, 
the escape-wheel being urged in the direction of the arrow. While this 
is occurring, the tooth b 2 encounters the projection n, and gives an 
impulse to the balance; the detent meanwhile has dropped back under 
the influence of the spring s, and catches the next tooth of the wheel 61. 

It will be noticed that the impulse is given to the balance immediately- 
after it has been subject to the resistance of unlocking the detent d, 
thus immediately compensating this resistance; also that the impulse 
is given at every alternate swing of the balance. 

The motion of the balance is so adjusted that the impulse is given 
through equal distances on each side of the middle of its swing. 



PROBLEMS 

1. Referring to Fig. 1, page 7, if OR is a crank 18 ins. long turning 90 r.p.m. 
find the displacement and acceleration of point T when 6 = 60 degrees. 

2. If the periphery speed of a drill 2 ins. in diameter is not to exceed 40 ft. per 
minute what is the maximum number of r.p.m. at which it may be run? 



3. (a) What is the angular speed of this 
pulley in radians per second? 

(6) What is the linear speed of a point M on 
the circumference in feet per minute ? 

(c) What is the ratio of the linear speed of 
P to that of Ml 



Peob. 3 

4. Assume that in turning steel shafting a cutting speed of 80 ft. per minute is 
allowable. How fast must a piece of 6-in. shaft turn to give this speed? 



5. The Pelton water wheel shown in the figure 
is driven by a jet of water having a velocity of 
90 feet per second. The linear speed of the 
center line of the buckets is 0.6 that of the 
water. What is the diameter in inches to the 
center line of the buckets if the wheel is fast to 
a shaft that turns 100 r.p.m.? 





Prob. 5 



6. This emery wheel is 6 ins. in diameter and makes 
2546 r.p.m. 

(a) What is its surface speed in ft. per minute? 

(b) If the wheel turns as indicated and the piece of cast 
iron D is fed as shown by the arrow at the rate of 30 ft. 
per minute, what is the cutting speed ? 

(c) If D is fed in the same direction and at the same 
speed as before and if the emery wheel turns 2546 r.p.m. 
but in a clockwise direction, what is the cutting speed? 

336 




PROBLEMS 



337 



7. A pulley makes 400 r.p.m. A point on its outer surface has a linear speed 
of 4000 ft. per minute. What is the diameter? 



.. 8.' The sketch shows a grindstone 
operated by a treadle. If w is taken 
as 3 and the average speed of the sur- 
face of the stone is 540 ft. per minute, 
through what distance does the point 
F on the treadle move in 15 minutes? 
Through how many radians does it 
move in that time? 

Note. — The point E is on the verti- 
cal center line YY when the treadle is 
in the highest and lowest positions. 




I Fulcrum 



Prob. 8 



On a revolving wheel are two points located on the same radial line. If the 
wheel has an angular speed of 1000 radians per minute and if the linear speed of 
one point is 3500 ins. per minute greater than the other, how far apart are the 
points ? 

10. The shaft of a centrifugal drying machine has an angular speed of 4398.24 
radians per minute. What is the linear speed of a point 2 ft. out from the center 
of the shaft in ft. per minute and how many r.p.m. does the dryer make? 

11. A bell-crank lever of the type shown in Fig. 17 has arms that are 8 and 10 
ins. long and make an angle of 60 degrees with each other. If the long arm is hori- 
zontal and has a 100-lb. weight suspended from its end, what vertical pull must be 
exerted at the end of the short arm to balance the 100-lb. weight? 

12. In a rocker of the type shown in Fig. 18 the line of motion of the end of one 
arm is vertical while the line of motion of the end of the other is 45 degrees with the 
horizontal. One arm is 5 ins. long and the motion of its end along the horizontal 
line is 2| ins., the motion of the end of the other arm is \\ ins. along the 45-degree 
line. If the ends of the arms of the bell crank swing equally either side of the 
lines of motion, locate graphically the fulcrum of the rocker and show the center 
lines of the arms when in their mid-positions. 

13. The main driving pulley of a broaching machine is 18 ins. in diameter and turns 
at a speed of 400 r.p.m. If the pulley is driven by a belt from a 10-H.P. motor 
developing its full rated power, what is the effective pull of the belt ? 

14. A shaft making 200 r.p.m. carries a pulley 36 ins. in diameter driven by a belt 
which transmits 5 horse power. What is the effective pull in the belt? What 
should be the diameter of the pulley if the effective pull is to be 50 lbs. ? 

15. A 6-in. belt transmits 8 horse power, when running over a pulley 20 ins. in 
diameter running 200 r.p.m. If the maximum allowable tension is 80 lbs. per inch 
of width and if the sum of the tensions is assumed to be constant, with what tension 
was the belt put on the pulleys? 



338 PROBLEMS ^y 

16. What is the width of a fabric belt which transmits 45 horse power when run- 
ning on a 30-in. pulley which turns 550 r.p.m. ? The belt was put on with a tension 
of 70 lbs. per inch, the maximum tension is not to exceed 95 lbs. per inch, neglecting 
centrifugal force, and it is assumed that the sum of the tensions is a constant. 

17. Determine the width of a single belt to carry 40 horse power when running 
on a pulley 48 ins. in diameter which turns 300 r.p.m. The maximum tension per 
inch of width for a single belt is 65 lbs., neglecting centrifugal force, and it is assumed 
that T\ = 2\ T%. Determine the width of this belt, using the millwrights' rule. 

18. A 3-in. belt is designed to stand a difference in tension of 50 lbs. per inch of 
width, neglecting centrifugal force. Find the least speed at which it can be driven 
in order to transmit 20 horse power. 

19. To what difference in tension on the two sides of a belt does the millwrights' 

rule correspond ? If the maximum tension for a single belt is 65 lbs. per inch of 

T 
width, and for a double belt 150 lbs. per inch of width, what is the value of ^ in a 

1 2 

belt calculated by the millwrights' rule? Answer both parts of the problem for 
both single and double belt, neglecting the effect of centrifugal force. 

20. Shaft A, turning 120 r.p.m., drives shaft B at speeds of 80, 120, 180 and 240 
r.p.m. by means of a pair of stepped pulleys and a crossed belt. The diameter of 
the largest step on the driving pulley is 18 ins. Calculate the diameters of all the 
steps of both pulleys. 

21. Two shafts, each carrying a four-step pulley, are to be connected by a crossed 
belt. The driving shaft is to turn 150 r.p.m. while the driven shaft is to turn 50, 
150, 250 and 600 r.p.m. The smallest step of the driver is 10 ins. in diameter. 
Find the diameters of all the steps. 

22. Solve the preceding problem if an open belt is used instead of a crossed 
belt and if the shafts are 30 ins. apart. 

23. Two shafts carrying five-step pulleys are to be connected by a crossed belt. 
The driver is to turn 150 r.p.m. while the follower is to have speeds of 50, 100, 150, 
200 and 250 r.p.m. If the smallest step on either pulley is 8 ins. in diameter, find 
the diameters of all the steps to two decimal places. 

24. The feed mechanism of an upright drill is operated by an open belt running 
on three-step pulleys. The driving shaft turns 150 r.p.m. while the driven shaft 
turns 150, 450 and 900 r.p.m., the two shafts being 15 ins. apart. If the largest 
diameter of the driver is 18 ins. find the diameters of all the steps. If the steps of 
these pulleys had been calculated for a crossed belt but an open belt had been used 
on them the belt would have been found too short to run on some of the steps. 
State approximately how much too short it would have been for the worst case. 

25. A lathe having a five-step pulley is driven by a belt (assumed to be crossed) 
from a pulley of the same size on the countershaft. The countershaft is to have a 
constant speed, and the lathe is to have speeds of 60 r.p.m. and 135 r.p.m. when the 
belt is on the steps either side of the center step. If the minimum speed is 40 r.p.m. 
and the smallest diameter 4 ins., find proper speed of countershaft, maximum speed 
of lathe, and diameter of all steps on the pulleys. 

26. Each of a pair of equal five-stepped pulleys has diameters 20, 17f, 15J, 12J, 
10J ins. Were these pulleys designed for an open belt or a crossed belt? If the 
driving shaft turns 500 r.p.m. calculate the five speeds of the driven shaft. 

27. In a pair of stepped pulleys the driver has diameters of 31.62, 25.5, 20.53, 10 
ins. The smallest diameter of the driven pulley is 7.91 ins. and its largest diam- 
eter 30 ins. Is the belt crossed or open? Calculate the distance between centers 
of the shafts and the other two diameters of the driven pulley. 



PROBLEMS 



339 



28. Two shafts are connected by a crossed belt running on a pair of speed cones. 
The driving shaft has a constant speed of 135 r.p.m. while the driven shaft is to 
have a range of speeds from 45 to 300 r.p.m., the speeds to increase in arithmetical 
progression as the belt is moved equal distances along the cones. The smallest 
diameter of the driving cone is 3 ins. Find the diameters of the cones at the ends 
and at two intermediate points. Plot the cones (i size) if their length is 24 ins. 

29. A rope drive composed of 15 ropes is to transmit 120 horse power. If the 
pitch diameter of one of the sheaves is 4 ft. and if its angular speed is 1500 radians 
per minute, what is the effective pull in each rope ? 

• 30. A rope drive (Multiple System) consisting of 15 ropes is transmitting 200 
horse power when the speed of the ropes is 1100 ft. per minute. The maximum 
tension per rope is 650 lbs. which is J the breaking strength of the rope (expressed 

T\ 
as, "a factor of safety of 4")- Find the ratio ■==-• Suppose that 3 of the ropes 

2 . Ti 

should break, the remaining ropes carrying the whole load. If the ratio 7=- stays 

1 2 
as before, what does the maximum tension become and what the factor of safety 
on the rope? 

3i. A motor running 500 r.p.m. transmits 3 horse power through a chain drive. 
The pitch diameter of the driving sprocket is 3 ins. What is the effective pull in 
the chain and what is the maximum tension neglecting centrifugal force? 

32. A chain drive is transmitting 3 horse power when the speed of the chain is 
900 ft. per minute. What is the effective pull in the chain? Suppose the speed of 
the chain be increased to 1200 ft. per minute, if the power transmitted remains the 
same as before, what is the effective pull? 

75 R.P. 



33. How many r.p.m. does B make? 



34. Angular speed of S = § of the ang- 
ular speed of T. Calculate and find gra- 
phically the diameters of cylinders to 
connect them: 

(a) When they turn as shown by the 
full arrows. 

(b) When they turn as shown by the 
dotted arrows. 




# 



^ 



rr. 



-24 ■ 



K 



Prob. 34 



35. A and B are rolling cylinders connecting 
the shafts S and T. C and E are cylinders fast 
to these shafts and slipping on each other at P. 
Find the diameters of C and E if the surface speed 
of E is twice that of C. 



c 


^ 


H 

E 


H 


1 H 


1 


U-H 


1 V 



-16' Dr 
Pbob. 35 



*\4"D. 



340 



PROBLEMS 



36. The shaft of a grindstone is mounted on 
rollers, as shown. If the circumference of the 
stone has a speed of 628.32 ft. per minute, find 
its r.p.m. If the angular speed of the rollers is 
209.44 radians per minute, find their diameter. 




Prob. 36 



I J 



37. How far from the axis of T will the 
center of the roller R be located if the 
angular speed of shaft S is three times as 
great as that of T ? 



T 



<~m i 



nmm.\ 

s 



Prob. 37 



38. A and B are two shafts at right angles, in the same 
vertical plane. C is a disk carried by a supporting yoke 
on a horizontal shaft arranged so that C is always in 
contact with the equal conoids on A and B. A turns at a 
constant speed of 60 r.p.m. What is the maximum speed 
of .B? What is the minimum speed of fi? What is the 
speed of B when the yoke supporting C has turned 30 
degrees from its present position? (Assume no slipping.) 




39. A turns 100 and B 150 
r.p.m. as shown and are connected 
by rolling cones. Calculate the 
apex angle of each cone. If the 
base of cone on A is 3 ins. from 
the vertex, calculate the diameters 
of both cones. Solve also graphi- 
cally 



\ 



Prob. 39 



^M 



^ 






Prob. 40 



40. Two shafts A and B are connected by rolling cones and turn as shown. A 
makes 300 r.p.m. while B makes 100 r.p.m. Calculate the apex angle of each cone 
and the diameter of each base if the base of cone B is 2 ins. from the vertex. Solve 
also graphically. 



PROBLEMS 



341 



41. Shaft S makes 180 r.p.m. and shaft T makes 60 r.p.m. Draw a pair of 
frustra of cones to connect them. Base of smaller cone 1 in. in diameter. Element 
of contact 1 in. long. 

(a) When the shafts turn as shown by the full arrows. 

(6) When they turn as shown by the dotted arrows. 



^ 



<• 



3H i — 



'% 



"4r 



Peob. 41 



Prob. 42 



42. Shafts A, B and C are connected by cones in external rolling contact so 
that the revolutions A:B:C = 3:2:4. If the diameter of cone B is 6 ins. 
draw in the three cones giving the diameters of cones A and C. (Show method 
clearly.) 

43. Given a 4-pitch gear of 24 teeth. The addendum equals the module, the clear- 
ance is to be i of the addendum and the back lash is to be 2 per cent of the circular 
pitch. Calculate the following, giving results to three decimal places: the pitch 
diameter, the diameter of the blank gear before cutting the teeth (addendum diam- 
eter), the depth of teeth, the backlash and the width of tooth and width of space. 



44. In the given position 

angular speed of A 

angular speed of B 

is equal to the angular speed ratio of what 

two rolling cylinders? Prove that this is so. 

(Make diagram full size.) 




Prob. 44 



45. A pinion 6-ins. in pitch diameter is to 
drive a rack. The flank of the rack is an arc 
of a circle 2|-ins. in radius with its center 1| ins. 
below pitch line of rack as shown. Find two 
points on the face of the pinion starting with 
points on given flank § and \ in. from the 
pitch point. Assuming an addendum of § in. 
on pinion show arc of recess on the rack's 
pitch line. 




Prob. 45 



342 



PROBLEMS 



46. Given this flank of a rack, a straight line at 15 
degrees with line of centers, find two points on the face 
of the pinion about \ and ^ in. away from the pitch fine, 
and show path of contact in recess if the pinion drives 
turning right-handed. 




Peob. 46 



47. A pinion 5 ins. in diameter is to drive 
an annular 12 ins. in diameter. The flanks 
of the annular's teeth are arcs of circles with 
2-in. radius located as shown. Find three 
points, as a, b, c, on the face of the pinion. 
Show the path of contact in recess if the 
pinion's addendum is 1 in. Find the arc of 
recess and the angles of recess, and the length 
of the acting flank. 




48. A pinion 10-ins. in diameter with radial flanks is to be driven;by a pinion 4-in. 
diameter also having radial flank teeth. 

1° Find the respective faces. 

2° If the addendums = \ in. show the path of contact; the angles of approach 
and recess; the maximum angles of obliquity in approach and recess; and the 
lengths of the acting flanks. 

3° Draw the true clearing curve which the 4-in. pinion's tooth would trace 



49. Diameter of pitch circle of pinion 
= 3| ins. Diameter of pitch circle of an- 
nular = 8 ins. The flank of the annular is 
to be the arc of a circle 1J ins. in radius 
as shown. Find two points on the face 
of the pinion starting with points on given 
flank f and f in. from the pitch point. 
Show path of contact in approach if annular 
drives left-handed. 




Prob. 49 



PROBLEMS 343 

60. Involute gears, 22| degrees obliquity, 1-pitch, addendum = f the module, 
clearance = f the module, no backlash. A pinion having 9 teeth, turning clockwise, 
is to drive a gear of 12 teeth. Indicate the path of contact, and angles of approach 
and recess for each gear, also give the ratio of the arc of action to the circular pitch. 
Draw two teeth on each gear, having a pair of teeth in contact at the pitch point. 
Indicate the acting flank of the teeth on each gear. 

51. Find the diameter and number of teeth of the smallest 3-pitch pinion with 
20 degrees obliquity, which would allow an arc of recess = arc of approach = the 
circular pitch, and draw its pitch and addendum circles. If the pinion drives a rack 
what is the greatest allowable addendum for the rack? 

52. Involute gears, 15 degrees obliquity; a 30-tooth pinion 2-pitch is to drive a 
rack. How long can the arc of approach be? Can the arc of recess equal the circu- 
lar pitch and why ? 

53. An involute gear with 21 teeth, 3-pitch, 15 degrees obliquity, has an adden- 
dum diameter of 1\ ins. Draw its base circle and pitch circle. Could two such 
gears properly be used to connect two shafts 1\ ins. apart? Give reason for answer. 

54. Involute gears, 2-pitch, 15 degrees obliquity. A 24-tooth pinion is to drive a 
32-tooth annular. The arc of approach to be equal to \\ the pitch and arc of recess 
to be equal to the circular pitch. Draw the addendum circles. Is each of these arcs 

_ possible; (explain clearly the steps by which this is determined). What isJheJjmijL. 
of the path of contact in approach and in recess and why ? 

55. Two cycloidal gears with 10 and 12 teeth respectively; 2-pitch; radial flanks 
on each. 

1° Draw the pitch circles and the describing circles and give their diameters. 

2° If the addendum = \ in. and the 10-tooth pinion drives turning right handed 
show the path of contact. 

3° How long is the arc of action in terms of the pitch? How long must it be to 
just give perfect action? 

56. A pinion with 6 teeth 1-pitch is to drive one with 8 teeth. Radial flanks on 8- 
tooth and the same size describing circle for the flanks of the 6-tooth. The arc 
of approach to be f of the pitch and the arc of recess to be f of the pitch. 

1° Find the maximum angles of obliquity in approach and in recess in degrees. 
2° Is the given arc of action possible ? 

57. Cycloidal Gears. Interchangeable Set. 3-pitch. Radial flanks on a 15- 
tooth gear. Addendum equals module. Clearance equals \ of the addendum. 
An 18-tooth pinion drives a 39-tooth annular. Show path of contact. How many 
teeth would there be in the smallest annular that would gear with the 18-tooth pinion ? 
Show path of contact in this case. 

58. In a f-pitch, interchangeable set of cycloidal gears with addendum the same 
on all gears, it is found that two 8-tooth pinions will give a path of contact 2 
inches long. Could one of the 8-tooth pinions properly drive a 7-tooth pinion of 
the set? 

59. A pin-gear 1-pitch with 8 pins is to be driven by a rack. The pins are to be 
one-half the pitch in diameter, and the addendum on the rack's teeth is \\ ins. 
Find the true path of contact. Also draw the teeth for the rack and the pins for 
the gear, assuming no back lash, and a clearance of 0.1 in. 

60. A 6-tooth 1-pitch cycloidal gear has its flank describing circle equal to the 
pitch circle. It is to drive another 6-tooth gear like itself. Draw the addendum 
circles so that the arcs of approach and recess are each equal to § of the pitch, and 
show the path of contact. What would be the shapes of the teeth ? 



344 



PROBLEMS 




a^n 



°\ 



wr 6 



Peob. 62 



61. Shaft A turns 120 r.p.m. in the di- 
rection shown and drives shaft B by means 
of an open belt running on the right-hand 
steps of the pulleys. Shaft C is driven 
from B by a pair of gears so that C turns 
3 times for every 2 turns of B. Gear D 
has 26 teeth while E has 78 teeth. Shaft 
F carries a bevel gear of 12 teeth which 
drives one of 120 teeth on shaft G. Shaft 
G also carries H, a 4-pitch, 16-tooth gear 
which is in mesh with a sliding rack. What 
is the speed of the rack in inches per minute 
and does it move to the right or left ? 

62. In a broaching machine, the shaft A carries a pulley 
24 ins. in diameter which is driven by a belt from a 12-in. 
pulley on the countershaft overhead, the latter turning 
150 r.p.m. The gears B and D have 12 teeth each, while 
C and E have 60 teeth. Gear E is fast to F, which has 
10 teeth and a circular pitch of 1.047 ins. and which 
engages with rack G to which is attached the broach. 
Find the speed with which the broach is drawn through 
the work in inches per minute. 

63. In a brick-making machine is 
found this train of gears. A motor 
carrying pulley E, which is 6 ins. in 
diameter, drives the machine. The 
wide-faced roller F, 12 ins. in diam- 
eter, drives a conveyor belt. If the 
motor runs at 1200 r.p.m. what is the 
speed of the conveyor belt in ft. per 
minute? (Neglect the thickness of 
the belt.) p ROB . 63 

64. In a crane, the chain barrel is driven by a motor on the spindle of which is 
keyed a pinion of 14 teeth. This gears with a wheel of 68 teeth keyed to the same 
spindle as a pinion of 12 teeth. The last wheel gears with a wheel of 50 teeth keyed 
to the same spindle as a wheel of 25 teeth, and the latter gears with a wheel of 54 
teeth keyed to the chain barrel spindle. Chain barrel is 16|-ins. in pitch diameter. 
Sketch the arrangement and find r.p.m. of motor when 20 ft. of chain are wound 
on drum per minute. 



65. Effective pull on the belt is 250 lbs. 
W = 7000 lbs. A is 23 ins. in diameter. B, 
D, and F each have 18 teeth. E = 63 teeth; 
G = 50 teeth. H is 20 inches in pitch diam- 
eter. If there is a loss of power of 60 per cent, 
how many teeth must there be in gear C? 
Which is the tight side of the belt? 




Belt 




Prob. 65 



PROBLEMS 



345 



66. Sketch shows side elevation of a 
molding machine. The stock is fed through 
rolls A to cutter C which is driven by a 
quarter turn belt as shown. Rolls A are 
4| ins. in diameter, the upper one only being 
power driven. If the cutter C is 6 ins. in 
diameter, find the feed of the stock per 
revolution of cutter and the relative speed 
of cutter and work. (Cutter is shown be- 
hind the work.) 



v ( go r 




1 V°J 




SVi D/aT^:^;-^ / 


/T~^ 



Prob. 66 



67. A is an annular gear having 77 teeth, 
driving pinion B having 12 teeth. Numbers 
of teeth on the other gears as are given in the 
figure. If A makes 15 r.p.m. find the rate of 
slip between the cylinders R and S in feet per 
second. 



R S 

h f— +— tf&H 



17 

■24 

Prob. 67 



68. The back gears of an engine lathe train are to give a reduction in the ratio 
of f . Arrange a train to give this, using no wheel of less than 15 teeth. First pair 
of gears 4-pitch, second pair 3-pitch. Make reduction of speed by the two pairs as 
nearly equal as possible. 



Rev. B 



find suitable numbers of teeth for the four 
No gear to 



' Rev. A 
gears of this train, having all of the same pitch 
have more than 75 teeth nor less than 10 teeth, 



Prob. 



70. Shaft S has a constant speed of 100 r.p.m. 
Gears F, G, and H form a unit free to slide, but 
not to turn on shaft Si. Si is to have speeds of 
20, 200 and 860 r.p.m. Gear H has 80 teeth. 
Find numbers of teeth on all gears if they are 
all of the same pitch, and if gears A and B are 
equal. The slowest speed of Si is when E and 
H are in mesh. 




346 



PROBLEMS 



71. The sketch shows a nest of spur gears, each pair being 
always in mesh and all gea^s of the same pitch. F, G, and H 
form a unit keyed to shaft B. C, D, and E are loose on shaft A, 
but may be locked to the shaft one at a time. Find the num- 
bers of teeth on all gears if the revolutions of B for one of A are 
to be I, H, and 2\ as C, D and E respectively are keyed to A. 
No gear to have less than 14 teeth. 



c 





E 




G 


H 



F 
Prob. 71 



A 



72. Both gears on both shafts A and B are fast. 
The three gears on shaft C are fast together but not 
to the shaft and may be moved along shaft C as a 
unit. Shaft A has a constant speed of 60 r.p.m. and 
speeds of B are to be 240, 60 and 15 r.p.m. Find 
suitable numbers of teeth in all gears if they are all 
4-pitch. 






I 



r* 



^G 



Prob. 72 

73. The value of a train is to be approximately 113, with an allowable variation 
of 1. Find the least number of pairs and sketch the train, indicating the numbers 

f teeth on each gear. No gear to have more than 50 teeth nor less than 12 teeth. 

87 T. 

74. A, B, and C are gears having teeth 
as shown. 

1° If A turns +4 and arm turns —3, 
find turns of B and C. 

2° If A is not to turn and B turns +30, 
find turns of the arm. 



75. Gear A is fixed. The 
arm turns about the shaft on 
which A is located. Find the 
number of teeth on gear C if 
it makes three times as many 
absolute turns as B does but 
in the opposite direction. 




Prob. 75 



PROBLEMS 



347 



76. In this roller bearing D represents 
the fixed bearing in which the rollers are 
supported while & is the shaft. Assuming 
that there is pure rolling contact between 
the shaft and the rollers, and between the 
rollers and D, find the ratio of speed at 
which roller cage revolves to speed at which 
shaft revolves. 



77. 1° If A turns +3 
and the arm —5, find the 
turns of B, C, and D. 

2° Suppose two idlers to 
be used between E and D, 
other conditions remaining 
as before, find the turns of 
D. 




Prob. 76 




Prob. 77 



78. If A turns +38 times, how 
many turns does the arm make ? 




^ Fixed 



Prob. 78 



348 



PROBLEMS 



79. If shaft A makes 20 turns in a positive 
direction, find the number of turns of shaft B 
and its direction. 






Arm 



60 T. nH 



A 

3 



T- 30 T. 



Prob. 79 



80. D is a fixed gear, A is fast to the 
shaft with the 54-toothed gear, and P is the 
arm of the epicyclic train. If A turns —20 
find the turns of the arm, then find the turns 
of B. 



2 





r. 


P 


3 



I 


L 




















— 
1 






5 




A D 
■ 
8 


r . 



Prob. 



81. In this hoisting mechanism, A is 
a fixed annular having 100 teeth. The 
two idle pinions B are carried by the arm 
of the epicyclic train, which also carries 
the drum, as shown. Gear C, which is 
fast to the crank, has 70 teeth. Diameter 
of drum is 5 ins., length of crank is 21 
ins., force applied at crank is 75 lbs. 
Find the teeth on the pinions B and 
weight lifted, neglecting friction. 




Prob. 81 



PROBLEMS 



349 



SOT. 



\. For 36 turns of D, find how 20 *■ 



many turns of F and in which direction ? 



•\ 



24 T. 



era 



T 

30 L 



30 T, 20 r.. 

3 N 



Peob. 82 



20 T. 

D 

a p 



83. For -3 turns of D, 29 *N 
find how many of F and in C 

which direction ? 



^40 T. 



30 T. 



L c-y 



w 



#100 T, 

V 

Pkob. 83 



\p30 T. 



84. In this train find the 
ratio of the diameters C and 
D, if 3 turns of A as shown 
are to cause the arm to turn 
11 times. Must a crossed 
or an open belt be used if 
the arm turns as shown? 



i 



E? 



I 



E. 

r 



y#:0T. 

r :ZJ 



P 



Prob. 84 



350 



PROBLEMS 



85. Let shaft S turn +3 times. Find 
the turns of the arm. 




75 T. 



Prob. 85 



86. If A is a shaft coupled 
to a dynamo making 2500 
r.p.m., how many revolutions £ 
per minute does B make? 



87. The gear B is fixed. For 31 
turns of A in direction shown, how 
many turns does C make and does 
it turn in the same direction as A 
or opposite? 




Prob. 87 



^™ 



PROBLEMS 



351 



88. In this epicyclic train let A 
turn +6, find the turns of B. 



J 90 T. 



30 T. 

^-200 T. 
30 T. 

9 



V I '\50T, 



Prob. 88 



89. The pulley P makes 40 r.p.m. in the direction shown. 
What must be the lead of the screw if nut A is to rise 3f ins. 
in 45 seconds? Is screw right handed or left handed? 



Ji it 




Prob. 89 



90. What pressure in lbs. on the square inch is exerted on 
a liquid below the piston by a force of 40 lbs. at the rim of 
the hand-wheel? The screw has f-in. lead and is double 
threaded. Allow 20 per cent lost in friction. 



r 


<-h"d- 

f 






f 


\ 




<I2D. 


\ 


^-, 






:1Pl1==3 





Prob. 90 



91. Find lead in inches of the screw if a force 
of If tons is to be exerted at W by a pull of 75 
pounds on the rope in the groove of the wheel D 
which has an effective diameter of 4| ft. (Allow 
45 per cent friction loss.) 




Prob. 91 



352 



PROBLEMS 






92. If driving pulley, D, makes 300 r.p.m. 
at what rate is the cross rail raised? 



VtLead R.H, 



93. B and C are two equal gears. They 
may have no axial motion. D and E are 
two gears, E being four times as large as D. 
Shaft A is a square shaft turning with B, but 
free to slide through it. The screw threaded 
through C has a lead of j in. left handed. 
How many turns of the handle are needed 
and which way (front side of D going down 
or up) to move the screw 2 ins. to the right? 



r Miiiimium. 



94. 20 turns of F are to raise W 5§ ins. Pi = 0.5 ins. 
lead R.H. P 2 is right handed. What is the lead of P 2 ? 
Which way must F turn as seen from above (right handed 
or left handed 1 ) ? (Two possible solutions.) 



95. Screw S has 10 threads per inch (single) right handed 
and is fixed. Nut A may slide but cannot turn. How many 
(single) threads per inch has screw Si if 46 turns of the hand- 
wheel in the direction shown lower A 0.66 ins.? Are threads 
on Si right handed or left handed? If the hand-wheel had a 
rim-radius of 7 ins. and Si had 8 threads per inch (single) right 
handed what force would be necessary at the rim to raise 
weight (PT) of 16,800 lbs.? Allow 60 per cent lost in friction. 




Prob. 95 



PROBLEMS 



353 



fl r ^ 



96. The hub H of the 200-tooth 
gear forms the nut for the screw S. 
The graduated wheel is fast to the 
shaft with the two pinions. How 
far will S move along its axis when 
W is turned through the angle rep- 
resented by one division ? In which 
direction will S move if W turns 
with the arrow? 



97. If the mechanism of Prob. 
96 were changed as shown in this 
figure, how far would S move and 
in which direction if W turns with 
the arrow, one division? 




42 77?.. per 



Peob. 97 



98. In this differential screw, A and C are 
broad-faced pinions which are fast to each other 
and are turned by the crank shown. H is a 
fixed nut, E is a left handed screw having f in. 
lead and F is a right handed screw having | in. 
lead. How far does E move for 24 turns of the 
crank and in which direction relative to the 
crank rotation? 




99. In a screw cutting train (Figs. 240a, 240b), the lead screw has a lead of \ in. 
left handed. Gear A has 30 teeth, while gear B has 40 teeth. Find suitable num- 
bers of teeth for the change gears C and D, using no gear of less than 16 teeth, if the 



354 



PROBLEMS 



lathe is to cut threads from 4 to 20 per inch, also to cut 1 1 § threads per inch. Arrange 
results in the form of a table, using the least number of gears and having as few 
teeth as possible. Give the number of different change gears that must be fur- 
nished with the lathe and also the total number of teeth that must be cut to make 
the set of change gears. Should N and M both be used in cutting a right handed 
thread, or only M ? 



100. In a worm and wheel let the worm be triple-threaded 
and the diameter of the drum be 14 ins. How many teeth 
must the wheel have if 30 turns of the worm are to move W 
20 ins.? If R = 16| ins., what must F be, if W equals 8800 
lbs. actually lifted, 40 per cent being the loss due to friction? 
If the handle is "pushed" to raise the weight, is the worm 
right handed or left-handed? (Take vr = ^.) 




101. Worm A is double-threaded and its worm wheel 
has 36 teeth. Worm B has a lead of f in. The pitch 
diameter of the drum for the weight is one foot. The 
force, F, at the end of a 16-in. handle on B is 20 lbs. and 
W is 25,344 lbs. If 60 per cent is lost in friction what is 
the diameter of worm wheel C? (Take ir = - 2 ? 2 --) 




Pkob. 101 



102. F is a double-threaded right handed worm. A is a worm wheel having 32 
teeth. On the same shaft with A is a gear C, 17-inch pitch diameter in mesh with 
gear D, 4-in. pitch diameter. On the shaft with D 
is the left-handed worm E having a lead of 2 ins., 
in mesh with worm wheel B, 10.83 ins. pitch diam. 
Disk H is fast to the shaft of worm F, and B is 
loose on this shaft. . 

1° How many turns of handle before H and B 
will be in the same position relative to one another ? 

2° What changes in these results would occur if 
worm E were right-handed instead of left-handed? 

3° If a drum 20 ins. in diameter were attached 
to B and a weight of 2000 lbs. suspended from it, 

how large a force at the handle would be necessary to raise the weight? 
friction. 




Neglect 



PROBLEMS 



355 



103. P is threaded through the worm wheel but can- 
not turn. Lead of thread on P is | in. right handed. 
Worm is double-threaded and right-handed. How many 
turns of B, and which way (right-handed or left-handed) 
to raise weighty in. ? What weight can be raised by a 
force of 50 lbs. applied at F? (30 per cent friction loss.) 




104. This plate cam is made up of arcs of two circles 
and their common tangents and turns about the fixed 
center A . Plot a curve showing the motion of the follower 
for every 15 degrees movement of the cam for \ turn. 
Scale of plot is to be as f ollows : Abscissae = angles turned 
by cam, f in. = 15 degrees. Ordinates = full size dis- 
placements of the follower. 

Make, on the same plot, a curve that would show 
the displacements of the follower, if its motion had been 
harmonic. 




Prob. 104 



105. The plate cam with axis at A consists of the arcs of 
three circles with centers and' radii as shown. Cam turns uni- 
formly left-handed. Draw a diagram which shall show the 
motion of the follower, ordinates to be distance moved by the 
follower (full size), and abscissse to represent angular motion of 
the cam (J in. = 30 degrees). Take points every 30 degrees 
with an extra point at 225 degrees. 




Prob. 105 



356 



PROBLEMS 



106. Find the outline of a plate cam which, by turning about 
the center C as shown by the arrow, shall cause a point A to 
move along the path A-B at a uniform speed as follows : T % the 
distance A-B in J turn of the cam. Still | turn. Remaining 
part of the distance in | turn. Return to A at once over the 
same path as previously traversed. Still \ turn. (Take 10 in- 
tervals in the distance A-B.) Cam to be full size. 




Prob. 106 



107. Starting from the position shown, the 
slide is to drop 2 ins. with harmonic motion 
during f of a turn, to rise at once 1 inch, to 
remain still § of a turn, to drop 2 ins. with 
uniformly accelerated and uniformly retarded 
motion in § turn, and then to rise 3 ins. at 
once. Find the cam outline if the end A of the 
lever is in contact with the cam, the latter to 
turn in the direction shown. (Assume that A 
is kept in contact with the cam by some ex- 
ternal force.) 




Prob. 107 



108. Find the outline of a plate cam turning 
uniformly right-handed to give block A the following 
motion: remain still T 1 s turn, rise 1 in. with harmonic 
motion in | turn, still £ turn, drop If ins. at once, 
rise f in. uniformly in \ turn. Cam is to drive roller 
B 1 in. in diameter. 




PROBLEMS 



357 



109. Piece A carries a pin which projects into 
the slot on the horizontal piece B. Find outline 
of a plate cam turning uniformly right-handed to 
act at D and give A the following motion: Still 
for \ turn of cam; up 1| ins. with harmonic 
motion in f turn; still \ turn, drop 1| ins. at 
once, and still \ turn. 




Peob. 109 



110. A and B are two rollers (f-in. diameter) attached to 
the same frame. The rollers are in the same plane and both 
are always to be in contact with a single plate cam. Find 
the outline of the cam if the frame is to be raised 1 in. with 
harmonic motion in \ turn of the cam. What will be the motion 
of the follower during the remaining \ turn of cam? 




Prob. 110 



111. Referring to Fig. 261, p. 211, a cam turning uniformly in a clockwise direc- 
tion, on the axis E, is to give the following motion to the follower S, the lowest 
position of the flat surface of S being 2 ins. above E. Up 2 ins. with harmonic 
motion in £ turn of the cam, down 1 in. with harmonic motion in \ turn, still J turn, 
down 1 in. with harmonic motion in J turn. Find the shape of the cam. 

112. Referring to Fig. 262, p. 212, the cam turns in a clockwise direction on 
axis C, the pivot P for the lever it! is 2\ ins. to the right and 3 ins. above C. The 
radius of the end of the slider S is f in. The center line of the slide S is 3 ins. to the 
left of C and when in its lowest position, its lowest point is 2 ins. above C. "T" is 
1 in., and a line parallel to the top surface of R and \ in. below it will pass through 
the pivot P. S moves up 3 ins. with uniformly accelerated and uniformly retarded 
motion in | turn of the cam, still £ turn, down 3 ins. with uniformly accelerated and 
uniformly retarded motion in § turn, still | turn. Find the shape of the cam. 



358 



PROBLEMS 



113. A cylindrical cam 3 ins. outside diameter (turning right handed as seen from 
the right) is to move a roller. Roller is above and moves parallel to the axis of the 
cam. Roller moves as follows: To right with harmonic motion 1| ins. in f turn of 
cam. Still for £ turn. To left with harmonic motion 1J ins. in i turn. Still for 
I turn. The roller is to be | in. in diameter at the large end and of such form as to 
give pure rolling contact. Groove in cam is to be f in. deep. Draw development 
for both top and bottom of groove of the part of cam which causes the motion to 
take place. 



114. A and D are fixed axes. AB = 1 
in., BC = If ins., BH = 3J ins., DC = 1| 
ins. Find the instantaneous axis of BCH. 
Assuming the velocity of B to be represented 
by a line f in. long, find the linear velocity 
of the point H on the rod BCH. 



115. A is a fixed axis. AB = 1 
in., BC = 2 ins., BH = 3 ins. Assum- 
ing the velocity of B to be represented 
by a line 1 in. long, find velocity of C 
and of H . 




Prob. 115 



116. A is a fixed axis. AB = 1 in., 
BC = 2 ins., BH = 3 ins., HE = 2 ins. 
Assuming velocity of B = 1 in., find 
velocity of E. 







Prob. 116 



117. A ladder 12 ft. long is leaning against a wall and makes an angle of 60 
degrees with the pavement. The wall is perpendicular to the pavement. If the 
lower end of the ladder slips on the pavement at the rate of 1 ft. per second, find 
graphically the velocity of the top of the ladder, the velocity of a point 4 ft. from 
the lower end and the velocity of the point on the ladder which is moving the 
slowest at this time. 



PROBLEMS 



359 




118. The velocity of A is repre- 
sented by a line 1| ins. long; find 
velocity of block P if the wheel rolls 
without slipping. 



Prob. 118 

119. The diagram represents the running gear of a locomotive. The driving 
wheels A, A are 6 ft. in diameter, while the trailing wheel B is 3 ft. in diameter. CD 
is 15 ins. and DE is 60 ins. The locomotive is moving forward with a velocity 
represented by a line 2 ins. long. Find graphically the absolute velocity of the 
crosshead E, and its velocity relative to the guides, also find the velocity of the top 
of the trailing wheel B. 




Prob. 119 



120. Velocity of A is represented 
by a line 1| ins. long. Find graphi- 
cally the velocity of B if B rolls with- 
out slipping on C. C and D turn to- 
gether and D rolls without slipping 
on surface E. 




Prob. 120 



121. A is a disk 2 ins. in diameter and B 
is a disk If ins. in diameter, both turning 
on fixed axes. A drives B with no shpping. 
C is a fixed axis. ED = 3f ins., CD = 1J 
ins., DX = If ins., FX = 3 ins. If the 
surface speed of A is 1 in., find the velocity 
of X. 




Prob. 121 



360 



PROBLEMS 



122. K, G, and H are fixed axes. The 
wheel is 2 ins. in diameter and rolls on 
track T without slipping. KM = 2±f 
ins., KA = l T 5 e ins., BD = If ins., GF = 1 
in., HE = | in., DC = 3| ins., DF = 2| 
ins., DE = l|f ins. If the velocity of A 
is | in., find velocity of C. 




Prob. 122 



123. Assuming no slip be- 
tween the cylindrical disks, and 
a surface velocity of 50 ft. per 
minute, find the velocity of the 
slide in ft. per minute. 




2" Ola 



124. Assuming no slip between the disks, and 
a surface velocity of disk A = 1 in., find the velocity 
of the center of disk X. 




Prob. 124 



iy 4 Dia, 



125. A, B and C are fixed cen- 
ters. Velocity of block P is repre- 
sented by a fine 1 J ins. long. Find 
velocity of P'. 




Prob. 125 



PROBLEMS 361 

126. If the velocity of the circumference of A = 1 in. find the velocity of X. 




Prob. 126 



127. D and E are fixed axes. DA = 3| ins., 
DF = If ins., EC = 2| ins., EG = 1A ins., GB = 3 
ins., F5 = 2 ins. If the velocity of A is 2 ins., find 
velocity of B. 




Prob. 127 



128. Given velocities of A 
and B represented by fines f 
and 1\ ins. long respectively. 
Find the velocity of S. 




Prob. 128 



362 



PROBLEMS 



129. Given the velocity of A repre- 
sented by a line 1 in. long, find the velocity 
of B. 




130. A 2-pitch pinion of 14 teeth is turning clockwise at the rate of 50 r.p.m. and 
is driving a rack. The tooth of the rack is a straight line making an angle of 22| 
degrees with the line of centers and the addendum of the pinion is equal to the 
module. Find by graphical method the rate of sliding between the teeth at the 
pitch point and at the end of recess. Give results in inches per minute, assuming 
7r = 3. Scale, one inch equals 300 ins. per minute. 



4*> 



131. In this linkage let BC = AB = CD = 1| 
ins., and AD = 2| ins. Find the centrode of BC 
between the positions BiCi and B%Ci and then find 
the centrode of AD if BC is fixed. 



132. AD (fixed) = 10 ins., BC = 10§ ins., 
DC = 4| ins. The sketch shows a linkage 
used in the feed mechanism of a vertical 
boring mill. The crank AB is adjustable 
for varying throws. Determine the angular 
motion of DC when AB is 2\ ins. (this being 
the setting for maximum feed). 




Prob. 131 




Prob. 132 



133. Plot a curve showing ratio 



angular speed CD 



angular speed AB 
starting with AB in position AB X and turning uni- 
formly left-handed. 

Ordinates = angular speed ratio (1 in. = unity) . 

Abscissae = angular positions of AB (| in. = 30 
degrees) . 

AC = \ in., DC = If ins., DB = 2\ ins., AB =2 



for 30-degree intervals of AB, 




Prob. 133 





PROBLEMS 363 

134. AC = BD = 8 ins., AB = CD = 3 ins. If 
^45 is turning uniformly 25 r.p.m. calculate the max- 
imum angular speed of CD in radians per minute. 
Sketch the linkage in the position at which the max- 
imum angular speed of. CD exists. 

Prob. 134 

135. A gas engine has a stroke of 5 ins. and a connecting rod 10 ins. long. 
Calculate, and check by graphical construction, the speed at which the piston is 
moving when it is at mid-stroke, if the engine is turning 1100 r.p.m. Calculate the 
acceleration of the piston when the crank is at a dead point. 

136. Disks M and N turn about their 

respective centers, C and B, within the frame 

K. Rod L slides in the [block at C which is 

fixed to disk M. The center line of the block 

coincides with a diameter of M . What is the 

, . angular speed disk M . . . 

value of ratio — — ; - — , ,. . — ^ in the 

angular speed disk N 

position shown? 

.„_...,,,.,, ,. . ,. time of cutting stroke 2 

137. A swmgmg-block quick return motion has a ratio -. ; t — ^— r — r~ = t' 

time or return stroke 1 

If the driving crank is 1 in. long, locate the fixed point of the swinging link. Draw 

the linkage in some convenient position, draw the infinite links which have been 

i j u 4-u vj- i. a j-u x- angular speed driven crank . ... 

replaced by the sliding pair, state the ratio — =_= — . . . . r I0r tnis 

^ angular speed drivmg crank 

position (numerical values not required) . State maximum (numerical) value of this 

ratio. 

138. In the swinging block mechanism, Fig. 335, let the maximum value of 

BA = 8 ins. and the minimum value = 3 ins.; the arm CN = 3 ft.; NP = 2 ft. 

2 ins.; path of P is perpendicular to CB and 19 ins. above B; maximum value of 

time of cutting stroke 2 . , nn 

-. j — 2— - — r— = T ; angular speed of gear M = 30 r.p.m. 

time of return stroke 1 ' ° * & ^ 

1° Find position of axis C. 

„ n _ , . . . j. time of cutting stroke 

2 Find minimum value ot -■ j 1 — 2 — r — r~ ' 

time of return stroke 

3° Plot a curve whose abscissae are time units and whose ordinates are linear 

speeds of P in feet per minute. Scale of abscissae \ in. = time occupied by gear M 

in turning through an angle of 15 degrees. Scale of ordinates 1 in. = 100 ft. per 

minute. 

139. In a Whitworth Quick Return mechanism similar to Fig. 340, BA = 5^ ins. 
BC = 12| ins. Assume that R is on a perpendicular to AB passing through A 
instead of above it, as shown in the figure. 

1° Find the ratio of time of cutting stroke to time of return stroke. 

2° Let AN = 15 inches; NR = 4 feet; angular speed of large gear = 1 radian 
per unit of time. Find the maximum linear speed of R in feet per unit of time. 

3° If the path of R is 20 ins. above A, other dimensions remaining the same 
as before, what difference results in ratio of time of cutting stroke to time of return 
stroke? 



364 



PROBLEMS 



140. Draw a pantograph to connect two points A and B, 1| ins. apart, so that 
the motion of A shall be to the motion of B as 13 is to 7. Calculate the distance 
from B to the fixed point. The pantograph is to be so arranged that A may move 
at least 5 ins. in either direction along the line through A and B. 

141. Design a pantograph to reduce the motion of the crosshead of an engine 
that has a stroke of 12 ins., to 3 ins., so that a steam engine indicator may be operated 
from it. Let the distance from the fixed point to the point of connection at the 
crosshead be 15 ins. Draw half size. 

142. The three points A, B, and C are to be connected by a pantograph so that 

A may move up 4 ins., B up 3 ins., and C down 2 ins. , , 

AC = 6 ins. Locate the fixed point and the point B, and T , 1 - 

then draw a pantograph that will allow A to be moved 4 B 

ins. in every direction. Prob. 142 



143. P is to have a stroke of 3 
ins. and is to lie on the same vertical 
fine in its mid-position (shown in 
sketch) and its two extreme posi- 
tions. Locate D and draw in the 
linkage. Then by producing AB 
and adding only two links, design a 
pantograph which shall contain a 
point whose motion is parallel to 
that of P and equal to f of it. Draw 
in the pantograph and designate 
the point clearly. 



\S 



V 




Prob. 143 



144. Stroke of P= 2 inches 
in the vertical straight line, 
locate the points C and B giv- 
ing the link BC and the moving 
point P. Then connect this 
point with a point R, 2 ins. 
horizontally to the right of P, 
by a pantograph so that 
motion of R _ 3 
motion of P 2 

and calculate the distance from 
R to the fixed point of the 
pantograph. 







< — 


— g» — 


H 






. 


V 


— <& 

D 




i % 


i 




A 








© 






B 





Prob. 144 



145. Referring to Fig. 374, the stroke of e is f of an inch downward from the 
position shown. A pencil at/ draws a line along ss 4 ins. long, the highest position 
of / being 2 ins. above line ca. ss is parallel to and 4f ins. to the right of tt. ca is 
perpendicular to tt and 1§ ins. above highest position of c. Link db is If ins. long. 
Link dc swings equal angles either side of a line through d parallel to tt. Locate d, 
find length of fc, dc, be and he; locate a and h. Solve by calculation or graphically, 
as is more convenient. Draw and dimension the mechanism. a 



PROBLEMS 



365 



-■ j"-'y7/77?/s 



146. Find W if there is a friction loss of 
40 per cent. F=75 lbs. 




r^n 



Pkob. 146 



147. In this hitch, what force F is required to raise a weight 
W of 1400 lbs., friction being neglected? 



rw\ 




148. If W = 3000 lbs., find the force F 




Prob. 148 



/ 149. Two men, weighing 150 lbs. each, stand on W and pull just 
enough to sustain the load. 

(a) "What pull do they exert on the rope? 

(b) What is the tension on the support for the upper block, 
neglecting the weight of the blocks and rope itself ? 

(c) If the men stood on the ground what would be the tension in 
the rope which supports the upper block? 




\W^I00Q*\ 
Pkob. 149 



366 PROBLEMS 

150. A differential pulley-block is to lift 1500 lbs. with a pull of 30 lbs., friction 
being neglected. Find the ratio of the larger diameter of the upper sheave to the 
smaller one. 

151. In a differential pulley-block, the smaller diameter of the upper sheave is 
12 ins. It is found necessary to haul over 7 ft. of chain to raise the weight 6 ins. 
What is the other diameter of the upper sheave? Neglecting friction, what weight 
would be raised by a pull of 40 lbs. 

152. With a differential pulley-block, if the diameters of the sheaves in the fixed 
block are 12 ins. and 11 ins., and if the weight of the lower block is 20 lbs., what net 
weight can be raised by a pull of 120 lbs. on the chain, allowing a loss of 30 per 
cent in friction ? How much chain must be overhauled to lift the weight one foot ? 



INDEX 



Acceleration, angular, 6. 

linear, 5. 

normal, 6. 

tangential, 6. 
Acting flank, 94. 
Action, angle of, 96. 

arc of, 96. 
Addendum, 94. 

circle, 94. 

limits of, on involute gears, 112. 
Aggregate combinations, 306. 

motion by linkwork, 306. 
Anchor escapement, 331. 
Angle of action, 96. 

of approach, 96. 

of recess, 96. 

pressure, 97. 
Angular acceleration, 6. 
Angular speed, 5. 
Annular gear, 86. 

involute, 117. 

cycloidal, 128. 
Approach, angle of, 96. 

arc of, 96. 
Arc of action, 96. 

of approach, 96. 

of recess, 96. 
Automobile differential, 173, 177. 

transmission, 160, 172. 
Axial pitch, 149. 
Axis, instantaneous, 228. 

Backlash, 95. 

Bands, 3. 

Barrel, 54. 

Bearings, 10. 

Bell crank lever, 16. 

Belt, approximate length of, 28. 

calculation of power, 25. 

connecting non-parallel shafts, 37. 

crossed, 23. 

double, 24. 



Belt drives, examples of, 40-49. 

exact length of, 29. 

open, 23. 

quarter-turn, 38. 

single, 24. 

tension in, 25. 
Belting, power of, 24. 
Belts, 21-50. 

kinds of, 23. 
Bevel epicyclic train, 176-179. 
Bevel gears, 86, 144-148. 

twisted, 86. 
Block chains, 57. 
Bush, 10. 

Cam and slotted sliding bar, 327. 
Cam, cylindrical, 197, 213-219. 

definition of, 197. 

diagram, 198-200. 

plate, 197, 200-212. 

plate, with flat follower, 210-212. 

positive motion plate, 209. 
Cams, combination of, 219-220. 
Centrode, 229. 

of a rolling body, 229. 
Chain, kinematic, 244. 
Chains, 21, 55-62. 

block, 57. 

calculation for length of, 59. 

conveyor, 55. 

hoisting, 55. 

Morse rocker-joint, 62. 

power transmision, 57 ■ 

Renold, 60. 

roller, 59. 

silent, 60. 
Chronometer escapement, 334. 
Chuck, elliptic, 283. 
Circular pitch, 95. 
Clearance, 94. 
Clearing curve, 104. 
Click, 311. 



367 



368 



INDEX 



Clockwork, 157. 

Closed pair, 9. 

Collar, 11. 

Combination, elementary, 1. 

Components, 225. 

Composition of velocities, 226. 

Compound screws, 188. 

Cones, rolling without slip, 67-72. 

Conic four-bar linkage, 286. 

Conjugate curves, 99. 

construction of, 99. 
Connecting rod, 221. 
Contact, path of, 97. 

pure rolling, 63. 
Continuous motion, 4. 
Conveyor chains, 55. 
Cords, 21, 50. 

small, 53. 
Cotton card train, 161. 
Counter mechanism, 322, 325. 
Coupling, Oldham's, 284. 
Crank, 16, 221. 

and rocker, 250. 

pin, 221. 
Crown gears, 86, 147. 
Crown-wheel escapement, 330. 
Crowning of pulleys, 49. 
Cycloidal gears, 122-138. 

low numbered teeth, 135. 
Cylinder and sphere, rolling, 73. 
Cylindrical cam, 197, 213-219. 

multiple turn, 215-218. 
Cylinders, rolling without slip, 63-66. 

Dead-beat escapement, 332. 

Dead points, 249. 

Diametral pitch, 95. 

Differential, automobile, 173, 177. 

Differential screws, 188. 

Direction, 3. 

Disk and roller, 74. 

Drag link, 252. 

Driven wheel, 153. 

Driver, 2. 

Driving wheel, 153. 

Drum, 54. 

Eccentric, 275. 

rod, 275. 
Eccentricity, 275. 



Effective lever arms, 20. 
Effective pull, 25. 
Elementary combination, 1. 
Elements, expansion of, 274. 

pairs of, 8. 
Ellipses, rolling, 81, 82. 
Elliptic chuck, 283. 

trammel, 281. 
Engine, oscillating, 270. 
Epicyclic train, 166-179. 
Epitrochoid, 104. 
Escapement, 311. 

anchor, 331. 

chronometer, 334. 

crown-wheel, 330. 

dead-beat, 332. 

Graham cylinder, 333. 
Escapements, 330. 
Expansion of elements, 274. 

Face, of gear, 94. 

of tooth, 94. 

width of, 94. 
Feather, 11. 

Ferguson's paradox, 171. 
Flank, acting, 94. 

of tooth, 94. 
Flexible connectors, 21. 
Follower, 2. 
Foot-pound, 24. 
Four-bar linkage, 221-245. 

angular speed ratio, 246. 

diagrams for change in speed ratio, 
248. 

non-parallel crank, 254. 

parallel crank, 254. 

relative motion of links, 246. 

sliding slot, 273. 

swinging block, 267. 

turning block, 271. 
Frame, 2. 

Frequency of contact, 156. 
Friction catch, 318. 
Friction gearing, 74-76. 

grooved, 76. 

Gear teeth, involute, 106-122. 

law governing, 98. 

rate of sliding of, 241. 
Gear train design, 162-165. 



^^™ 



INDEX 



369 



Gears, annular, 86. 

automobile differential, 173, 177. 

bevel, 144-148. 

classified, 86. 

crown, 86, 147. 

cycloidal, 122-138. 

cycloidal, low numbers of teeth, 135. 

drives, 86. 

epicyclic, 166-179. 

face of, 94. 

helical, 86, 149-152. 

helical, speed ratio, 152. 

herringbone, 86, 139. 

hyperboloidal, 86. 

interchangeable, 119. 

involute, standard proportions of, 122. 

mitre, 86. 

pin, 139-143. 

reduction, 162. 

screw, 148. 

separation of, 119. 

skew, 86. 

skew bevel, 147. 

sliding eliminated, 139. 

stepped, 138. 

twisted, 138. 

twisted bevel, 147. 
Geneva stop, 325. 
Graham cylinder escapement, 333. 
Gudgeon, 10. 
Guide pulleys, 40. 
Guides, 10. 

Harmonic motion, 7. 
Helical gears, 86, 149-152. 

speed ratio, 152. 
Helix, the, 148. 
Herringbone gears, 86, 139. 
Higher pair, 9. 
Hookes' joint, 287-290. 
Horsepower, 24. 

Horsepower of belt, calculation of, 25. 
Hoisting chains, 55. 
Hoisting machine train, 162. 
Hunting cog, 156. 
Hyperbolas, rolling of, 83. 
Hyperboloidal gears, 86. 



Instantaneous axis, 228. 

of a rolling body, 229. 
Instantaneous center, 229. 
Interchangeable, involute gears, 119. 
Intermittent motion, 4. 

from continuous motion, 324. 

from reciprocating motion, 311. 
Inversion of pairs, 10. 
Involute, applied to gear teeth, 106. 
Involute gears, separation of, 119. 
Involute, of a circle, 105. 
Isosceles linkage, 278-284. 

Journal, 10. 

Keys, 11, 
Keyway, 11. 
Kinematic chain, 244. 

Lead, 149. 

Lever, double rocking, 264. 

nipping, 319. 
Lever arms, effective, 20. 
Levers, 16-20. 

bell crank, 16. 

kind of, 16. 

motion from, 17. 
Line of centers, 221. 
Line of connection, 22. 
Linear acceleration, 5. 

speed, 4. 
Link, 3, 222, 244. 
Linkage, 244. 

angular speed ratio, 246. 

conic, 286. 

four-bar, 221, 245. 

isosceles, 278-284. 

relative motion of links, 246. 

sliding block, 259. 

sliding slot, 273. 

turning block, 271. 
Links, relative motion of, 224. 
Linkwork, slow motion by, 257. 
Linkwork with a sliding pair, 258. 
Locking devices, 328. 
Logarithmic spirals, rolling, 78-81. 
Lower pair, 9. 



Idle wheel, 154. 
Inclined plane, 180, 181. 



Machine, 1. 
Mashed wheels, 321. 



370 



INDEX 



Mechanism, 1. 

constructive, 2. 

pure, 2. 
Mitre gears, 86. 
Module, 95. 

Morse rocker-joint chain, 62. 
Motion, 3. 

continuous, 4. 

cycle of, 4. 

from levers, 17. 

graphical representation of, 224. 

harmonic, 7. 

intermittent, 4. 

kinds of, 7. 

modification of, 8. 

parallelogram of, 225. 

parallelopiped of, 226. 

period of, 4. 

reciprocating, 4. 

resultant, 225. 
Multiple turn cylindrical cam, 215-218. 

Neck, 10. 

Nipping lever, 319. 

Non-cylindrical surfaces, rolling of, 77- 

85. 
Normal acceleration, 6. 
Normal pitch, 109, 149. 

relation to circular pitch, 110. 
Nut, 10. 



Obliquity of action, 97. 
Oldham's coupling, 284. 
Oscillating engine, 270. 
Oscillation, 4. 



Pawl, 312. 

double acting, 315. 

reversible, 313. 
Peaucellier's straight line mechanism, 

291. 
Pedestal, 10. 
Periphery-speed, 14. 
Pillow-block, 10. 
Pin gears, 139-143. 
Pitch, axial, 149. 

circles, 94. 

circular, 95. 

diametral, 95. 

normal, 109, 149. 

number, 95. 

point, 94. 

surface, 22. 
Pivot, 10. 

Plane, inclined, 180, 181. 
Planer drive, 158. 
Plate cam, 197, 200-212. 

with flat follower, 210. 
Plumber-block, 10. 
Positive motion plate cam, 209. 
Power transmission chains, 57. 
Power, unit of, 24. 
Pressure angle, 97 
Pulley, 21. 
Pulley-block, triplex, 175. 

Weston differential, 310. 
Pulley-blocks, 307. 
Pulleys, crowning of, 49. 

guide, 40. 

stepped, 30 

stepped, equal, 34. 

tight and loose, 50. 



Pairs, lower, 9. 

higher, 9. 

incomplete, 9. 

inversion of, 10. 

of elements, 8. 
Pantograph, 298. 
Parabolas, rolling, 83. 
Parallel motion by cords, 305. 

by means of four-bar linkage, 303. 
ParaUelogram of motion, 225. 
Parallelopiped of motion, 226. 
Path, 3. 
Path of contact, 97. 



Quarter-turn belt, 38. 
Quick return, swinging block, 267-270. 
Whitworth, 271. 

Rack and pinion, 86. 

involute, 116. 
Radian, 12. 
Ratchet wheel, 311. 
Recess, angle of, 96. 

arc of, 96. 
Reciprocating motion, 4. 
Reduction gear, 162. 
Renold silent chain, 60. 



INDEX 



371 



Resolution and composition of velocities, 

typical problems, 230-243. 
Resolution of velocities, 226. 
Resultant, 225. 
Resultant motion, 225. 
Revolution, 4. 
Revolving bodies, 12. 

angular speed of, 12. 

linear speed of, 12. 
Robert's straight-line mechanism, 302. 
Rocker, 16. 
Roller chains, 59. 

Rolling bodies, velocities of point on, 230. 
Rolling contact, pure, 63. 
Root, circles, 94. 

distance, 94. 
Rope driving, 51-55. 

systems of, 51. 

grooves for, 53. 
Ropes and cords, 50. 
Ropes, wire, 54. 

wire, grooves for, 55. 
Rotation, 4. 

axis of, 4. 

direction of, 4. 

plane of, 4. 



Speed, relation between forces and, 
15. 

surface, 14. 

uniform, 5. 

variable, 5. 
Spindle, 10. 
Spline, 11. 

Sprockets, diameters of, 60. 
Spur gears, 86. 

twisted, 86. 
Star wheel, 327. 
Step, 10. 
Stepped pulleys, 30. 

for crowned belt, 31. 

for open belt, 33. 
Stepped wheels, 138. 
Straight-line mechanism, 291. 

Peaucellier's, 291. 

Robert's, 302. 

Scott Russell, 292-294. 

Tchebicheff's, 303. 

Watt's, 295-298. 
Sun and planet wheel, 170. 
Surface speed, 14. 
Swash-plate, 285. 
Swinging-block linkage, 267. 



Scott Russell, straight-line mechanism, 

292-294. 
Screw, 10. 

cutting, 190-193. 

gears, 148. 

rotation caused by axial pressure, 190. 

threads, 181. 
Screws, differential, 188. 

speed ratio, 187. 
Shafts connected by belt, directional re- 
lation, 22. 

speed ratio, 22. 
Silent chains, 60. 
Silent feed, 320. 
Skew bevel, 147. 

gears, 86. 
Slides, 10. 

Sliding-block linkage, 259-266. 
Sliding slot linkage, 273. 
Speed, angular, 5. 

cones, 36. 

linear, 4. 

periphery, 14. 



Tangential acceleration, 6. 
Tchebicheff's straight hne mechanism, 

303. 
Threads, left-hand, 186. 

per inch, 186. 

right-hand, 186. 

screw, 181-187. 
Tight and loose pulleys, 50. 
Train, automobile transmission, 160. 

bevel epicyclic, 176-179. 

cotton card, 161. 

design of, 162-165. 

epicyclic, 166-179. 

hoisting, 162. 

value, 154. 
Trains of wheels, 153. 
Trammel, elliptic, 281. 
Turn, 4, 12. 

Turning block linkage, 271. 
Transmission, automobile, 160, 172. 

modes of, 2. 
Triple pulley block, 175. 
Twisted bevel gears, 86, 147. 



372 



INDEX 



Twisted spur gears, 138. 
Uniform speed, 5. 
Universal joint, 288. 

Velocities, composition and resolution of, 
226. 
graphical representation of, 224. 
of rigidly connected points, 226. 
typical problems, 230-243. 



Velocity, 5. 
Vibration, 4. 

Watt's straight-line mechanism, 295-298. 

Wedge, 180-181. 

Whitworth quick return, 271. 

Wiper, 197. 

Wire ropes, 54. 

Worm and wheel, 86, 193-196. 



D 



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%m 



